Binary Search Problem |LeetCode Problem – 1|Easy

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Count Negative Numbers in a Sorted Matrix       

Problem – Given a m * n matrix grid which is sorted in non-increasing order both row-wise and column-wise.  Return the number of negative numbers in grid using Binary search in this problem

Example 1:

Input: grid = [[4,3,2,-1],[3,2,1,-1],[1,1,-1,-2],[-1,-1,-2,-3]]

Output: 8

Explanation: There are 8 negatives number in the matrix.

Example 2:

Input: grid = [[3,2],[1,0]]

Output: 0

Example 3:

Input: grid = [[1,-1],[-1,-1]]

Output: 3

Example 4:

Input: grid = [[-1]]

Output: 1


  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 100
  • -100 <= grid[i][j] <= 100


According to size, if we go in linear fashion, that is, Check each cell of matrix one by one the time complexity is O (m*n)  which is affordbale if m and n less then 100

But if we try to do it in O(m*log(n)) time, We have to use Binary Search, which is better approach.

Linear Search Solution

Algorithm –

Simply traverse one by one each cell of matrix and increase count if the value is negative

int countNegatives(vector<vector<int>>& grid) {
        int ans = 0;
        for(int i =0;i<grid.size();i++)
            for(int j = 0;j<grid[i].size();j++)
        return ans;
Time Complexity – O(no of rows* no of column)

Binary Search Solution

As we know, Binary search is a divide and conquer algorithm, so, we are going to focus on dividing each row of the array in this problem

  1. Traverse each row one by one
  2. Do a Binary search to find the first negative element index
  3. Subtract the length of row by index to find the total number of a negative number, because elements are arranged in descending order
  4. Repeat Until any row left
class Solution {
    int countNegatives(vector<vector<int>>& grid) {
        int ans = 0;
        for(int i =0;i<grid.size();i++)
            ans+=binarySearch(grid[i]); // Traversing each row one by one
        return ans;
    int binarySearch(vector<int> &bs)
        int  l = 0;
        int  r = bs.size()-1;
            int mid = (l+r)/2;
            {                    // for finding negative element
                r = mid-1;
                l = mid+1;
        if(l<bs.size() && bs[l]<0)
        return bs.size()-l;   // return number of negative element
            return 0;
Time Complexity  = O(no of rows * log( max of column ))       

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