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Prefix Sum of Matrix (Or 2D Array) in C++
In this problem, we are given a 2D array of integer values mat[][]. Our task is to print the prefix sum matrix of mat.
Prefix sum matrix: every element of the matrix is the sum elements above and left of it. i.e
prefixSum[i][j] = mat[i][j] + mat[i-1][j]...mat[0][j] + mat[i][j-1] +... mat[i][0].
Let’s take an example to understand the problem
Input: arr =[ [4 6 1] [5 7 2] [3 8 9] ] Output:[ [4 10 11] [9 22 25] [12 33 45] ]
To solve this problem, one simple solution is finding prefixSum by traversing all elements till i,j position and add them. But it is a bit complex for the system.
A more effective solution will be using the formula for finding the values of elements of prefixSum matrix.
The general formula for element at ij position is
prefixSum[i][j] = prefixSum[i-1][j] + prefixSum[i][j-1] - prefixSum[i-1][j-1] + a[i][j]
Some specail cases are
For i = j = 0, prefixSum[i][j] = a[i][j] For i = 0 and j > 0, prefixSum[i][j] = prefixSum[i][j-1] + a[i][j] For i > 0 and j = 0, prefixSum[i][j] = prefixSum[i-1][j] + a[i][j]
The code to show the implementation of our solution
Example
#include <iostream>
using namespace std;
#define R 3
#define C 3
void printPrefixSum(int a[][C]) {
int prefixSum[R][C];
prefixSum[0][0] = a[0][0];
for (int i = 1; i < C; i++)
prefixSum[0][i] = prefixSum[0][i - 1] + a[0][i];
for (int i = 0; i < R; i++)
prefixSum[i][0] = prefixSum[i - 1][0] + a[i][0];
for (int i = 1; i < R; i++) {
for (int j = 1; j < C; j++)
prefixSum[i][j]=prefixSum[i- 1][j]+prefixSum[i][j- 1]-prefixSum[i- 1][j- 1]+a[i][j];
}
for (int i = 0; i < R; i++) {
for (int j = 0; j < C; j++)
cout<<prefixSum[i][j]<<"\t";
cout<<endl;
}
}
int main() {
int mat[R][C] = {
{ 1, 2, 3},
{ 4, 5, 6},
{ 7, 8, 9}
};
cout<<"The prefix Sum Matrix is :\n";
printPrefixSum(mat);
return 0;
}Output
The prefix Sum Matrix is : 1 3 6 5 12 21 12 27 45
Updated on: 04-Feb-2020
2K+ Views
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