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. Constraints n, m ≤ 250 where n and m are the number of rows and columns in matrix matrix[i][j] ≤ 2**12 Example 1 Input matrix = [ [2, 3], [5, 7] ] Output [ [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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