Matrix Prefix Sum - Amazon Top Interview Questions

Problem Statement :

Given a two-dimensional integer matrix, return a new matrix A of the same dimensions where each element is set to A[i][j] = sum(matrix[r][c]) for all r ≤ i, c ≤ j.


n, m ≤ 250 where n and m are the number of rows and columns in matrix
matrix[i][j] ≤ 2**12

Example 1

matrix = [
    [2, 3],
    [5, 7]

    [2, 5],
    [7, 17]

Solution :


                        Solution in C++ :

vector<vector<int>> solve(vector<vector<int>>& matrix) {
    for (int i = 0; i < matrix.size(); ++i) {
        for (int j = 0; j < matrix[0].size(); ++j) {
            if (i - 1 >= 0) matrix[i][j] += matrix[i - 1][j];

            if (j - 1 >= 0) matrix[i][j] += matrix[i][j - 1];

            if (i - 1 >= 0 && j - 1 >= 0) matrix[i][j] -= matrix[i - 1][j - 1];

    return matrix;

                        Solution in Java :

import java.util.*;

class Solution {
    public int[][] solve(int[][] matrix) {
        if (matrix.length == 0 || matrix[0].length == 0) {
            return matrix;

        int[][] output = new int[matrix.length][matrix[0].length];
        output[0][0] = matrix[0][0];
        for (int x = 0; x < matrix.length; x += 1) {
            for (int y = 0; y < matrix[x].length; y += 1) {
                int sum = 0;
                for (int a = 0; a <= x; a += 1) {
                    for (int b = 0; b <= y; b += 1) {
                        sum += matrix[a][b];
                output[x][y] = sum;
        return output;

                        Solution in Python : 
class Solution:
    def solve(self, matrix):

        if not matrix:
            return []

        r = len(matrix)
        c = len(matrix[0])

        for i in range(r):
            for j in range(1, c):
                matrix[i][j] += matrix[i][j - 1]

        for j in range(c):
            for i in range(1, r):
                matrix[i][j] += matrix[i - 1][j]

        return matrix

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