- nx is the biggest sale event of the year, when many products are heavily discounted.
- Since its widespread popularity, differing theories have spread about the origin of the name "Black Friday."
- The name was coined back in the late 1860s when a major stock market crashed.

C For Loop: Exercise-2 with Solution. **Write** a C program **to find** the **sum** of **first** 10 natural **numbers**. Pictorial Presentation: Sample Solution: C Code:. All you need to do is look for two numbers that multiply to give 42. There are several possible answers: 1 and 42 2 and 21 3 and 14 6 and 7 -1 and -42 -2 and -21 -3 and -14 -6 and -7 What are the product and sum of 2, 4 and 9? Work out the product of 2, 4 and 9. The product means that you need to multiply the three numbers together. 2 × 4 × 9 = 72. Related tutorials How to get the **first** element of an array in C++ How to get the **first** character of a string in C++ C++ program to add two **numbers** How to remove the **first** character of a string in C++ Checking if a string is empty in C++ How to **write** comments in C++ How to iterate over the words of a string in C++ How to get the last element of an array in C++ How to Convert String to Double in. **Sum** All **Odd** Fibonacci **Numbers** Problem Explanation You will need to gather all the Fibonacci **numbers** and then **check** for the **odd** ones. Once you get the **odd** ones then you will add them all. The last **number** should be the **number** given as a parameter if it actually happens to be an off Fibonacci **number**. Relevant Links Fibonacci **number** Hints Hint 1 To get the next. * // License: GPL * // * // Changes: See changelog.txt * // ***** define('mPDF_VERSION', '6.1'); //Scale factor define('_MPDFK', (72 / 25.4)); // Specify which font. . **To find average** of N **numbers** with max **numbers** and its values are given from user. Approach : Read an input integer for asking max **numbers** using input() or raw_input(). Jun 24, 2020 · **Sum of first** 5 natural **numbers** is 15. In the above program, the **sum** of the **first** n natural **numbers** is calculated using the formula. Then this value is displayed. This is demonstrated by the following code snippet. **sum** = n* (n+1)/2; cout<<"**Sum of first** "<<n<<" natural **numbers** is "<<**sum**;. Logic **to find sum** of natural **numbers** using. The **sum** of the **first odd** integers , beginning with one, is a perfect square: 1, 1 + 3, 1 + 3 + 5, 1 + 3 + 5 + 7, etc. This explains Galileo's law of **odd numbers** : if a body falling from rest covers one unit of distance in the **first** arbitrary time interval, it covers 3, 5, 7, etc., units of distance in subsequent time intervals of the same length. **Algorithm** **to** **find** the **sum** **of** **odd** **numbers**. Step1: Take a variable named N and store the value of the upper limit of N. Step2: Take one more variable named **sum** **to** store the result of the calculation. Step3: Take a for loop with variable i which be initialized with 1 and in condition expression check for (i<=N) and. What is the **Algorithm** to print the all even **numbers** from 1 to 100? Two solutions immediately spring to mind:Run over all the **numbers** in the range and if the **number** is even. What is the **Algorithm** to print the all even **numbers** from 1 to 100? Two solutions immediately spring to mind:Run over all the **numbers** in the range and if the **number** is even. If s < 7 (mod 8) the **number** can be expressed as a **sum** of three squares. Subtract a square and **check** whether the result has the form 2 m (4k+1) where m is a non-negative integer. If not, select another square and repeat. Finally **check** if the **number** 4k+1 is prime. If not, select another square and repeat. Jun 13, 2022 · Difference between sums of **odd** and even digits; **Find the sum of digits** of a **number at even and odd places**; Count even and **odd** digits in an Integer; Count number of even and **odd** elements in an array; Program to print **Sum** of even and **odd** elements in an array; **Sum** of even **numbers** at even position; Even **numbers** at even index and **odd** **numbers** at **odd** .... Input upper limit to **find** **sum** **of** **odd** **numbers** from user. Store it in some variable say N. Initialize other variable to store **sum** say **sum** = 0. To **find** **sum** **of** **odd** **numbers** we must iterate through all **odd** **numbers** between 1 to n. Run a loop from 1 to N, increment 1 in each iteration. The loop structure must look similar to for (i=1; i<=N; i++). Jun 24, 2020 · **Sum of first** 5 natural **numbers** is 15. In the above program, the **sum** of the **first** n natural **numbers** is calculated using the formula. Then this value is displayed. This is demonstrated by the following code snippet. **sum** = n* (n+1)/2; cout<<"**Sum of first** "<<n<<" natural **numbers** is "<<**sum**;. Logic **to find sum** of natural **numbers** using. **Find** if a given number “n” is **odd** or even. A number is even if it can be divided by 2 without remainder. Such **numbers** are 2, 4, 6, 8.. and so on. The **numbers** that leave a remainder are called **odd**. They are 1, 3, 5, 7.. and so on. In programming we **find** the remainder of a division with the operator %.. Calculate and print: (i) The average of the total marks obtained by N number of students. [average = (sum of total marks of all the students)/N] (ii) Deviation of each student's total marks with the average. [deviation = total marks of a student - average] View Answer Bookmark Now.

**Algorithm** **to** **find** the **sum** **of** **odd** **numbers**. Step1: Take a variable named N and store the value of the upper limit of N. Step2: Take one more variable named **sum** **to** store the result of the calculation. Step3: Take a for loop with variable i which be initialized with 1 and in condition expression check for (i<=N) and. The program **first** asks the user to input two values P and Q for a range (where p < Q). After that, the program should **find** a **number** having highest **number** of divisors. In the end, the program should print that **numbers** (having highest **number** of divisors) along its divisor count. Sample Inputs: Enter **first number** (P): 1 Enter second **number** (Q): 10.

Here is a list of Egyptian fractions for 1 arranged in order of denominator **sum** up to **50**: Denoms of EF for 1 Denom **sum**; 1: 1: 2, 3, 6 : 11 : 2, 4, 6, 12 ... it uses just the oddness and evenness of **numbers**. Any **odd number** can be **written** as 2n + 1 and every even **number** is of ... Is it possible **to find** a **sum** of **odd** Egyptian fractions. The concepts are as follows: 1) Matlab functions 2) if else in matlab 3) rem function in matlab (remainder after division) 4) fprintf in matlab Program function isevenorodd (n) %This function will check whether the given number is even or odd and %will print even if the number is even and odd if the number is odd.

## mt

In this problem, we need **to find the largest number with** a given **number** of digits N and given **sum** of digits say M. We will use a greedy **algorithm** to solve this is O (N) time complexity. Example: If N (**number** of digits) = 3 and M (**sum** of digits) = 15. Then, the largest **number** satisfying it will be 960. There are other **numbers** as well but they. The **sum** of the **first** n **odd** natural **numbers** is (2k-1 represents any **odd number**): [6.1] We can expand the left-hand side: [6.2] And use our formula for the **sum** of the natural **numbers**: [6.3] Rounding up like terms, the **sum** of the **first** n **odd** natural **numbers** is: [6.4] Application - **Sum** of Even **Numbers** The **first** even **numbers** are (2k is always even. However, it only contains n - 1 distinct **numbers**. So one of the **numbers** is missing and another **number** is duplicated. **Write** a Java method that takes A as an input argument. Repeat the following steps while N > 0. Set r = N % 10. Store the rightmost digit of N in r. Set N = N / 10. Remove the rightmost digit of N. If r is even, set **sum** = **sum** + r. If r is **odd**, do nothing. **The sum of even digits** of N is stored in **sum**. #**include**<stdio.h>. The program output is also shown below. /* * C program to find the sum of first 50 natural numbers * using for loop */ #include <stdio.h> void main () { int num, sum = 0; for ( num = 1; num <= 50; num ++) { sum = sum + num; } printf("Sum = %4d\n", sum); } Program Explanation. Place the below line of code after LINE B to print the average of numbers. int average = sum/input; System.out.println("Average of numbers: " + average); Change LINE A to replace the code with the below code so that it prints the sum of squares. sum = sum + (i*i); Try it Online Factorial Program In Java Using for Loop. 1) Write a program to calculate the area of triangle using formula at=√s (s-a) (s-b) (s-c) Click here for solution 2) Basic salary of an employee is input through the keyboard. The DA is 25% of the basic salary while the HRA is 15% of the basic salary. Provident Fund is deducted at the rate of 10% of the gross salary (BS+DA+HRA). Suppose a sequence of **numbers** is arithmetic (that is, it increases or decreases by a constant amount each term), and you want **to find** the **sum** of the **first** n terms. Denote this partial **sum** by S n . Then. S n = n ( a 1 + a n) 2 , where n is the **number** of terms, a 1 is the **first** term and a n is the last term. The **sum** of the **first** n terms of an. **Algorithm to find sum** of integers in the string. Step1: Start. Step2: Take string as input from the user. ... This Python **sum** of **odd numbers** program is the same as above. But, we used the third parameter inside the for loop to eliminate the If block. ... Example: How **to find** the **sum** of even and **odd** elements of an array in Python import array as. . Answer (1 of 3): A simple loop will do it. i = 1; while i is equal or less than **50** { if is_**odd**(i) then print i; i = i + 1 } You need to **write** a function **to find** is i is **odd**. For example: is_**odd** (i) {. (**1st number** in set + last **number** in set)/2 Remember, we must average the **first** even integer in the set and the last even integer in the set. So we have: (100 + 300)/2 = 400/2 = 200 Next we have to determine the quantity. Once again, we **include** the **first** even integer in the set and the last even integer in the set. STEP 4: Add a and b , and store in **SUM**. STEP 5: Display the value of **SUM**. STEP 6. Stop. Flowchart. Greatest among Two **numbers**. **Algorithm**. STEP 1: Start. STEP 2: Read Two **numbers** a and b. STEP 3: Compare a and b. If a is greater than b then print a otherwise print b. STEP 4: Stop. Flowchart . **First** ten natural **numbers**. **Algorithm**. STEP 1: Start.

Procedure **to find** the **sum** of even and **odd numbers** in Java array , a) Take an array . b) Take two variables to store the **sum** of even and **odd numbers**. Assume they are evenSum, and oddSum. Initialize them with 0. c) Traverse the array . d) **Check** each element of the array .. . **Algorithm**: **sum** (n) 1) **Find number** of digits minus one in n. To find the sum of even numbers we need to iterate the numbers from 1 to n. Flowchart Output N=10 Sum of even number from 1 to N =30 In the same way, we can calculate the sum of odd numbers too from the 1 to N range. To calculate the sum of odd numbers, instead of the even number 2 we need to start with the odd number 1. sum = 0 for i in range(1, 101): sum = sum + i print(sum) Output: 5050 Python Code Editor Online - Click to Expand Python code implementation using the function In this code, we will print the sum of the first 100 Natural Numbers using the function. Code: def sum_100_natural_numbers(): sum = 0 for i in range(1, 101): sum = sum + i return sum. So for i th element in the array will have appearances at the **first** position in all the sub-arrays will be = (n-i). So for the **first** position, occurrences are. 1 appears 4 times. 2 appears 3 times. 3 appears 2 times. 4 appears 1 time. From Step 1 if we subtract the **number** of occurrences in above step, the remaining occurrences are (i is the. **First** give a meaningful name to our function, say sumOfEvenOdd (). Next the function accepts two integer values from user i.e. start and end range. Hence, update function declaration to sumOfEvenOdd (int start, int end);. Finally, after calculating **sum** **of** even or **odd** **numbers** the function must return it to the caller. STEP 4: Add a and b , and store in **SUM**. STEP 5: Display the value of **SUM**. STEP 6. Stop. Flowchart. Greatest among Two **numbers**. **Algorithm**. STEP 1: Start. STEP 2: Read Two **numbers** a and b. STEP 3: Compare a and b. If a is greater than b then print a otherwise print b. STEP 4: Stop. Flowchart . **First** ten natural **numbers**. **Algorithm**. STEP 1: Start. Algorithm STEP 1: Start STEP 2: COUNT = 0 STEP 3: Read the value of number in N STEP 4: COUNT = COUNT + 1 STEP 5: Compute N * COUNT ; STEP 6: Is count < = 10 {Yes:. Output Enter a positive integer: 20 **Sum** = 210 Suppose the user entered 20. Initially, addNumbers () is called from main () with 20 passed as an argument. The **number** 20 is added to the result of addNumbers (19). In the next function call from addNumbers () to addNumbers (), 19 is passed which is added to the result of addNumbers (18). **To** store add we are using AL register. Initially set AL to 0. To check the **number** is even or **odd**, we have used TEST 01 instruction. This instruction performs ANDing with the data and 01H. So if the LSB is 0, the total result will be 0, otherwise it will be 1. When LSB is 1, it indicates that the **number** is **odd**, then add them into one. The sum of even numbers are: 90 Methodology: First, define an array with elements. Next, declare and initialize two variables to find sum as oddSum=0, evenSum=0 Then, use the “while loop” to take the elements one by one from the array. The “if statement” finds a number and then if the number is even, it is added to evenSum. User entered value for this Java Program to find Sum of Odd Numbers : number = 5 For Loop First Iteration: for (i = 1; i <= 5; i++) if (i % 2 != 0) => if (1 % 2 != 0) – Condition is True. oddSum. **Write an algorithm** to add given two **numbers**. User input validation **to check** a **number**. Storing New Line character (\n) as a user input data. Draw **flowchart** to display Good morning message bas... Draw **flowchart to check** negative **number**. Draw **flowchart to check** Positive **Number**. Draw **Flowchart to check Odd** or Even **Number**. April (3). I am looking to create **an algorithm** that produces all of the **numbers** that **sum** to a **number** under certain rules. There should be exactly two different natural **numbers** that are. It is constructed by starting with ones at the outside and then always adding two adjacent **numbers** and **writing** the **sum** directly underneath. ... (2x)2 = 4x2 which is even. Since an **odd number** can be expressed as 2x + 1, (2x + 1)2 = 4x2 + 4x + 1. 4x2 and 4x are even, ... The **first 50** Fibonacci **numbers** Fn for n = 0, 1, 2, ... ,49 are:[12]. **Write** **an** **Algorithm** **to** **find** the **sum** **of** **first** ten even **numbers**. Also make its Flowchart. Get the answers you need, now! ... answered • expert verified **Write** **an** **Algorithm** **to** **find** the **sum** **of** **first** ten even **numbers**. Also make its Flowchart. 2 See answers Advertisement Advertisement ... // and half **odd** **numbers**.. The formula of arithmetic progression is So we can conclude that the sum of first n odd numbers is square of n. C Program to Find Sum of first n odd numbers #include <stdio.h> int main(void) { int n,sum; printf("Enter the number"); scanf("%d",&n); sum = n*n; printf("Sum of first %d odd numbers are %d",n,sum); return 0; } Output :. Ex . 6 : **Find** the **sum** of the cubes of the **first** 25 **odd numbers**. Sol: **First** 25 **odd** cube **numbers** means 13 + 33+ 53 + ———-+493 So Here n = 25. = 25 2 [ (2 x 252 )– 1 ] = 625 x [ 1250 – 1] =625 x 1249 = 780625. Ex . 7 : **Find** the **sum** of the consecutive cube **numbers** 263+283+ 303 + 323—–+1003 . Sol : 263+283+ 303 + 323—–+1003 = {23. User entered value for this Java Program to find Sum of Odd Numbers : number = 5 For Loop First Iteration: for (i = 1; i <= 5; i++) if (i % 2 != 0) => if (1 % 2 != 0) – Condition is True. oddSum. Here is a list of Egyptian fractions for 1 arranged in order of denominator **sum** up to **50**: Denoms of EF for 1 Denom **sum**; 1: 1: 2, 3, 6 : 11 : 2, 4, 6, 12 ... it uses just the oddness and evenness of **numbers**. Any **odd number** can be **written** as 2n + 1 and every even **number** is of ... Is it possible **to find** a **sum** of **odd** Egyptian fractions.

There are two ways to proceed: step through the integers, testing each for **odd**; or start from a known **first** **odd** **number** and repeatedly construct the next **odd** **number**. The latter approach might be considered cheating ... depending on the precise rules (or constraints) on the assignment. Jan 21 '13 # 3 reply swapnali143 34. You are mixing everything into a single **algorithm**. In most cases it's a better idea to have separate steps. In your case, the steps could be: Count **odds** and evens **Find** the outlier Print the result That way you can make sure that the output does not interfere with the loops. **Find** if a given number “n” is **odd** or even. A number is even if it can be divided by 2 without remainder. Such **numbers** are 2, 4, 6, 8.. and so on. The **numbers** that leave a remainder are called **odd**. They are 1, 3, 5, 7.. and so on. In programming we **find** the remainder of a division with the operator %.. C For Loop: Exercise-2 with Solution. **Write** a C program **to find** the **sum** of **first** 10 natural **numbers**. Pictorial Presentation: Sample Solution: C Code:. Calculate and print: (i) The average of the total marks obtained by N number of students. [average = (sum of total marks of all the students)/N] (ii) Deviation of each student's total marks with the average. [deviation = total marks of a student - average] View Answer Bookmark Now.

step 3 : accept second **number** step 4 : add these two **numbers** step 5 : display result step 6 : stop //**write an algorithm to find** the **sum** of three **numbers**. step 1 : start step 2 : accept all three **numbers** step 3 : add all three **numbers** and store in one variable step 4 : display the result step 5 : stop //**write an algorithm to find** the.

## aw

**To find average** of N **numbers** with max **numbers** and its values are given from user. Approach : Read an input integer for asking max **numbers** using input() or raw_input(). **The sum of two numbers is 12**. Experiment with a situation in which **the sum of two numbers is 12**. **1st**: You enter the **1st number**. Enter negatives as "-x" rather than "- x." 2nd: Compute the other **number** mentally yourself. 3rd: Press the button in each row to display the other **number**. 4th: Look for a pattern: **Find** a method for computing the other. **Find** the largest palindrome made from the product of two 3-digit **numbers**. **To check** if a **number** in Javascript is palindrome, we can convert it to String, then split into char array, reverse the array, and join as a string, then a palindrome is a string that its reverse is the same. You are mixing everything into a single **algorithm**. In most cases it's a better idea to have separate steps. In your case, the steps could be: Count **odds** and evens **Find** the outlier Print the result That way you can make sure that the output does not interfere with the loops. **Algorithm** and Flowchart **to find** the **Sum** of **first** n **odd numbers**(1,3,5....)using the equationn x n. The **sum** **of** the **first** natural **number** is 1. **Sum** **of** **first** two natural **numbers** is 1 + 3 = 4 = 2*2. **Sum** **of** **first** three natural **numbers** is 1 + 3 + 5 = 9 = 3*3. **Sum** **of** **first** four natural **numbers** is 16 = 4*4. Hence proved, the **sum** **of** **odd** natural **numbers** is given by n2 where n is the **number** **of** **odd** terms that you are going to add. 3. The sum of the first n Fibonacci numbers with odd indices is f1 + f3 +...+ f2n-1 = f2n. The proof is similar to the proof in II. Example: f1 + f3 + f5 = 1+ 2 + 5 = 8 = f6 . IV. The sum of the first n Fibonacci numbers with even indices is f2 + f4 + ... + f2n = f2n+1 - 1. Proof: From II, f1 + f2 + f3 + ... + f2n = f2n+2 - 1. Procedure **to find** the **sum** of even and **odd numbers** in Java array , a) Take an array . b) Take two variables to store the **sum** of even and **odd numbers**. Assume they are evenSum, and oddSum. Initialize them with 0. c) Traverse the array . d) **Check** each element of the array .. . **Algorithm**: **sum** (n) 1) **Find number** of digits minus one in n. Sum from a to n = [n * (n + 1) / 2] – [ (a - 1) * a / 2] We want to get rid of every number from 1 up to a – 1. How about even numbers, like 2 + 4 + 6 + 8 + + n? Just double the regular formula. To add evens from 2 to 50, find 1 + 2 + 3 + 4 + 25 and double it:. The program **first** asks the user to input two values P and Q for a range (where p < Q). After that, the program should **find** a **number** having highest **number** of divisors. In the end, the program should print that **numbers** (having highest **number** of divisors) along its divisor count. Sample Inputs: Enter **first number** (P): 1 Enter second **number** (Q): 10. # Take input from user. num = int(input("Print sum of odd numbers till : ")) sum = 0; for i in range(1, num + 1): #Check for odd or not. if(not (i % 2) == 0): sum += i; print("\nSum of odd.

## jd

What is FCFS (**first** come **first** serve)disk scheduling:-. FCFS is the simplest of all the Disk Scheduling **Algorithms**. In FCFS, the requests are addressed in the order they arrive in the disk queue. Example: Given the following queue -- 95, 180, 34, 119, 11, 123, 62, 64 with the Read-**write** head initially at the track **50** and the tail track being at. 4. Work any of your defined formulas **to find** the **sum** . Once you've plugged in the integer, multiply the integer by itself plus 1 , 2 , or 4 depending on your formula. Then divide your result by 2 or 4 to get the answer. For the example of consecutive formula 100∗101/2, multiply 100 by. Procedure **to find** the **sum** of even and **odd numbers** in Java array , a) Take an array . b) Take two variables to store the **sum** of even and **odd numbers**. Assume they are evenSum, and oddSum. Initialize them with 0. c) Traverse the array . d) **Check** each element of the array .. . **Algorithm**: **sum** (n) 1) **Find number** of digits minus one in n. How to **Write** PseudoCode to **Find** **Sum** **of** Natural **Numbers** (1-100) PseudoCode: [crayon-631f3dc6746a0278208041/]. **Sum** and average of n **numbers** in Python. Accept the **number** n from a user. Use input() function to accept integer **number** from a user.. Run a loop till the entered **number**.. How to **Write** PseudoCode to **Find** **Sum** **of** Natural **Numbers** (1-100) PseudoCode: [crayon-631f3dc6746a0278208041/]. Answer (1 of 3): A simple loop will do it. i = 1; while i is equal or less than **50** { if is_**odd**(i) then print i; i = i + 1 } You need to **write** a function **to find** is i is **odd**. For example: is_**odd** (i) {. Nov 16, 2020 · When any **number** which ends with 0,2,4,6,8 is divided by 2 that is an even **number**.And when any **number** ends with 1,3,5,7,9 is not divided by two is an **odd number**.Example: Input : 8 Output: **Sum of First** 8 Even **numbers** = 72 **Sum of First** 8 **Odd numbers** = 64 Approach #1: Iterative. Jan 10, 2019 · In java program to print **odd** and even. Jun 24, 2020 · **Sum of first** 5 natural **numbers** is 15. In the above program, the **sum** of the **first** n natural **numbers** is calculated using the formula. Then this value is displayed. This is demonstrated by the following code snippet. **sum** = n* (n+1)/2; cout<<"**Sum of first** "<<n<<" natural **numbers** is "<<**sum**;. Logic **to find sum** of natural **numbers** using. Given a **number** n, **find** **sum** **of** square of **first** n **odd** natural **numbers**. Input : 3 Output : 35 1 2 + 3 2 + 5 2 = 35 Input : 8 Output : 680 1 2 + 3 2 + 5 2 + 7 2 + 9 2 + 11 2 + 13 2 + 15 2. Try It! A simple solution is to traverse through n **odd** **numbers** and **find** the **sum** **of** square. **C++ Programming Tutorial**. 1. Getting Started - **Write** our **First** Hello-world C++ Program. Let us begin by **writing** our **first** C++ program that prints the message "hello, world" on the display console. Step 1: **Write** the Source Code: Enter the following source codes using a programming text editor (such as NotePad++ for Windows or gedit for UNIX. **Algorithm**. We shall use the following **algorithm to find** the average of **numbers**. Start. Read array of **numbers**. Or initialize an array with **numbers**, of whom you would like **to find** average. Initialize **sum** = 0; For each **number** in the array, add the **number** to **sum**. Compute average = **sum** / **number** of elements in array. Stop. The original computes the **sum** **of** the **first** n **odd** **numbers**. Your **algorithm** computes the **sum** **of** all the **odd** **numbers** in the range 1..n. So for an input of n=3, the **first** **algorithm** will compute 1+3+5, while your **algorithm** will compute 1+3. (If you want a quicker way, then the formula n*n computes the **sum** **of** the **first** n **odd** **numbers**). Answer (1 of 32): 1.)Start 2.)Declare a variable of type int i=1 and sum=0 3.)Iteration for i=1 and i<=50 { **sum** = **sum** + i increment i = i + 1 } 4.)Print **Sum** 5.)Stop C++ Syntax [code]#include<iostream.h> using namespace std int main() { int i,sum=0; for(i=1;i<=50;++i) { **sum** += i;. **Sum** **of** **odd** **numbers** from 1 to **50** is : 625 Previous Python program to print all **odd** **numbers** from 1 to N Next Python program to print all even **numbers** from 1 to N Program tags program programming python programs. 4. Push into the Sub-arrays(blue rectangle) all the elements in the Previous step Added Sub-arrays (green rectangle) adding the element to each of them. 5. Copy the Recently Added Sub-arrays elements into the Previous Step Added Sub-arrays. 6. Repeat. When an element gets added to Sub-arrays (the blue rectangle), we can easily compute the **sum** and.