Quiz on Prefix Sum for DSA

This quiz will test your understanding of the Prefix Sum technique and how to efficiently compute cumulative sums and solve array-related problems using this powerful approach in data structures and algorithms.

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Question 1

What is a prefix sum array?


  • An array where each element is the sum of all elements to the right of it.

  • An array that stores the product of elements up to the current index.

  • An array that stores the sum of elements from the start to the current index.

  • An array where each element is the sum of all elements to the left of it.

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Question 2

In what scenarios is the prefix sum technique particularly useful?

  • When performing a merge sort

  • When the array has duplicate elements

  • When you need to compute the sum of elements in a range frequently

  • When you need to find the minimum element in an array

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Question 3

What will be the output of the following program :

C++ 
int main() {
    int arr[] = {4, -1, 2, 1};
    int prefix_sum[4];
    prefix_sum[0] = arr[0];
    for (int i = 1; i < arr.size(); i++) {
        prefix_sum[i] = prefix_sum[i-1] + arr[i];
    }
    for (int i = 0; i < arr.size(); i++) {
        cout << prefix_sum[i] << " ";
    }
    return 0;
}
C 
int main() {
    int arr[] = {4, -1, 2, 1};
    int prefix_sum[4];
    prefix_sum[0] = arr[0];
    for (int i = 1; i < 4; i++) {
        prefix_sum[i] = prefix_sum[i-1] + arr[i];
    }
    for (int i = 0; i < 4; i++) {
        printf("%d ", prefix_sum[i]);
    }
    return 0;
}
Java 
public class Main {
    public static void main(String[] args) {
        int[] arr = {4, -1, 2, 1};
        int[] prefix_sum = new int[arr.length];
        prefix_sum[0] = arr[0];
        for (int i = 1; i < arr.length; i++) {
            prefix_sum[i] = prefix_sum[i - 1] + arr[i];
        }
        // Printing the prefix sum
        for (int i = 0; i < arr.length; i++) {
        System.out.println(prefix_sum[i]); 
    }
    }
}
Python 
arr = [4, -1, 2, 1]
prefix_sum = [0] * len(arr)
prefix_sum[0] = arr[0]
for i in range(1, len(arr)):
    prefix_sum[i] = prefix_sum[i-1] + arr[i]
for i in range(len(arr)):
    print(prefix_sum[i], end=' ')
JavaScript 
let arr = [4, -1, 2, 1];
let prefix_sum = new Array(arr.length);
prefix_sum[0] = arr[0];
for (let i = 1; i < arr.length; i++) {
    prefix_sum[i] = prefix_sum[i - 1] + arr[i];
}
for (let i = 0; i < arr.length; i++) {
    console.log(prefix_sum[i] + ' ');
}


  • 4 3 5 6

  • 4 3 7 8

  • 3 2 1 5

  • 5 3 7 9

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Question 4

What is the time complexity of constructing a prefix sum array for an array of size n?


  • O(n)

  • O(log n)

  • O(1)

  • O(n^2)

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Question 5

What is the time complexity of querying the sum of elements in a subarray after constructing the prefix sum array?


  • O(n)

  • O(n^2)

  • O(log n)

  • O(1)

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Question 6

What happens if we modify the values of the original array after constructing a prefix sum array?

  • The prefix sum array will automatically update

  • The prefix sum array will keep the old values

  • The prefix sum array will update only the first element

  • The prefix sum array becomes invalid and needs to be recalculated

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Question 7

What would be the output if pass arr[][] = [[10, 20, 30], [5, 10, 20], [2, 4, 6]] to the following code:

C++ 
int[][] prefixSum2D(int a[R][C]) 
{ 
    int psa[R][C]; 
    psa[0][0] = a[0][0];
    for (int i = 1; i < C; i++) 
        {
            psa[0][i] = psa[0][i - 1] + a[0][i]; 
        }
    for (int i = 1; i < R; i++)
        {
            psa[i][0] = psa[i - 1][0] + a[i][0];
        }
    for (int i = 1; i < R; i++) 
        { 
        for (int j = 1; j < C; j++)
            {   
                psa[i][j] = psa[i - 1][j] + psa[i][j - 1] - psa[i - 1][j - 1] + a[i][j]; 
            }
        } 
    return psa; 
}
C 
int[][] prefixSum2D(int a[R][C]) { 
    int psa[R][C]; 
    psa[0][0] = a[0][0];
    
    for (int i = 1; i < C; i++) {
        psa[0][i] = psa[0][i - 1] + a[0][i]; 
    }

    for (int i = 1; i < R; i++) {
        psa[i][0] = psa[i - 1][0] + a[i][0];
    }

    for (int i = 1; i < R; i++) { 
        for (int j = 1; j < C; j++) {   
            psa[i][j] = psa[i - 1][j] + psa[i][j - 1] - psa[i - 1][j - 1] + a[i][j]; 
        }
    }
    return psa; 
}
Java 
public static int[][] prefixSum2D(int[][] a) {
    int[][] psa = new int[R][C];
    psa[0][0] = a[0][0];
        
    for (int i = 1; i < C; i++) {
        psa[0][i] = psa[0][i - 1] + a[0][i]; 
    }

    for (int i = 1; i < R; i++) {
        psa[i][0] = psa[i - 1][0] + a[i][0];
        }

        for (int i = 1; i < R; i++) { 
            for (int j = 1; j < C; j++) {   
                psa[i][j] = psa[i - 1][j] + psa[i][j - 1] - psa[i - 1][j - 1] + a[i][j]; 
            }
        }
    }
    return psa; 
}
Python 
def prefix_sum_2d(a):
    psa = [[0] * C for _ in range(R)]
    psa[0][0] = a[0][0]

    for i in range(1, C):
        psa[0][i] = psa[0][i - 1] + a[0][i]

    for i in range(1, R):
        psa[i][0] = psa[i - 1][0] + a[i][0]

    for i in range(1, R):
        for j in range(1, C):
            psa[i][j] = psa[i - 1][j] + psa[i][j - 1] - psa[i - 1][j - 1] + a[i][j]
            
    return psa;
JavaScript 
function prefixSum2D(a) {
    let psa = Array.from({ length: R }, () => Array(C).fill(0));
    psa[0][0] = a[0][0];

    for (let i = 1; i < C; i++) {
        psa[0][i] = psa[0][i - 1] + a[0][i];
    }

    for (let i = 1; i < R; i++) {
        psa[i][0] = psa[i - 1][0] + a[i][0];
    }

    for (let i = 1; i < R; i++) {
        for (let j = 1; j < C; j++) {
            psa[i][j] = psa[i - 1][j] + psa[i][j - 1] - psa[i - 1][j - 1] + a[i][j];
        }
    }
    return psa; 
}

  • [[10, 30, 60],[5, 15, 35],[2, 6, 10]]

  • [[10, 20, 30], [15, 30, 50], [17, 34, 56]]

  • [[10, 30, 60], [15, 45, 95], [17, 51, 107]]

  • None of the above

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Question 8

How do you calculate the sum of elements in a subarray using the prefix sum array?

  • By directly summing the elements of the subarray

  • By multiplying the prefix sum values at the left and right boundaries

  • By subtracting the prefix sum value of the left boundary index from the prefix sum value of the right boundary index

  • By dividing the prefix sum values at the left and right boundaries

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