Skip to main content

Prefix Sum of 2D matrix code

 Today we will see the code to get Prefix sum matrix of a 2D matrix .


We calculate the prefix sum by changing the matrix in-place.

For the 0th row and column, prefix sum can be easily calculated by adding the previous sum to the element.
For the other cells, we can find the prefix sum as:

  1. Add the prefix sum of the cell just above and left to it.
  2. Subtract the prefix sum of the cell on the top-left diagonal cell. We do this because we have added this prefix sum in each of the top and left cell of the given cell. Therefore, we have to subtract it once.


Fig.Getting prefix sum






Code : 

for(int i=0; i<row; i++){
            for(int j=0; j<col; j++){
                if(i>0) grid[i][j] += grid[i-1][j];
                if(j>0) grid[i][j] += grid[i][j-1];
                if(i>0 && j>0) grid[i][j] -= grid[i-1][j-1];

               
            }
        }

Related question for practice : 


https://leetcode.com/problems/count-submatrices-with-top-left-element-and-sum-less-than-k/description/


Labels:

Comments

Post a Comment

Popular posts from this blog

leetcode 48 solution

  48 .  Rotate Image You are given an  n x n  2D  matrix  representing an image, rotate the image by  90  degrees (clockwise). You have to rotate the image  in-place , which means you have to modify the input 2D matrix directly.  DO NOT  allocate another 2D matrix and do the rotation.   Example 1: Input: matrix = [[1,2,3],[4,5,6],[7,8,9]] Output: [[7,4,1],[8,5,2],[9,6,3]] Example 2: Input: matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]] Output: [[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]   Constraints: n == matrix.length == matrix[i].length 1 <= n <= 20 -1000 <= matrix[i][j] <= 1000 solution: class Solution { public:     void swap(int& a , int &b)     {         int c ;         c = a;         a = b;         b = c;     }     void transpose (vector<vector<int>...
Post a Comment
Read more

Regular Expression Matching Leetcode Solution

Regular Expression Matching Given an input string s and a pattern p, implement regular expression matching with support for '.' and '*' where: '.' Matches any single character.​​​​ '*' Matches zero or more of the preceding element. The matching should cover the entire input string (not partial). Example 1: Input: s = "aa", p = "a"  Output: false  Explanation: "a" does not match the entire string "aa". Example 2: Input: s = "aa", p = "a*"  Output: true  Explanation: '*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa". Example 3: Input: s = "ab", p = ".*"  Output: true  Explanation: ".*" means "zero or more (*) of any character (.)". Constraints: 1 <= s.length <= 20 1 <= p.length <= 20 s contains only lowercase English letters. p contains only lowercase Englis...
Post a Comment
Read more

2485. Find the Pivot Integer | Binary search

  Given a positive integer   n , find the   pivot integer   x   such that: The sum of all elements between  1  and  x  inclusively equals the sum of all elements between  x  and  n  inclusively. Return  the pivot integer  x . If no such integer exists, return  -1 . It is guaranteed that there will be at most one pivot index for the given input.   Example 1: Input: n = 8 Output: 6 Explanation: 6 is the pivot integer since: 1 + 2 + 3 + 4 + 5 + 6 = 6 + 7 + 8 = 21. Example 2: Input: n = 1 Output: 1 Explanation: 1 is the pivot integer since: 1 = 1. Example 3: Input: n = 4 Output: -1 Explanation: It can be proved that no such integer exist.   Constraints: 1 <= n <= 1000 Solution : class Solution { publ ic:     int pivotInteger( int n ) {         int sum = (( n )*( n + 1 ))/ 2 ;         int i = 1 ;         int j =...
Post a Comment
Read more