378. Kth Smallest Element in a Sorted Matrix

Problem Statement

Given an n x n matrix where each of the rows and columns is sorted in ascending order, return the kth smallest element in the matrix.

Note that it is the kth smallest element in the sorted order, not the kth distinct element.

You must find a solution with a memory complexity better than O(n2).

Example 1:

Input: matrix = [[1,5,9],[10,11,13],[12,13,15]], k = 8
Output: 13
Explanation: The elements in the matrix are [1,5,9,10,11,12,13,13,15], and the 8th smallest number is 13

Example 2:

Input: matrix = [[-5]], k = 1
Output: -5

Constraints:

• n == matrix.length == matrix[i].length

• 1 <= n <= 300

• -109 <= matrix[i][j] <= 109

• All the rows and columns of matrix are guaranteed to be sorted in non-decreasing order.

• 1 <= k <= n2

Intuition

We use Upperbound as sorted to find min no of elements before 
For every mid

Links

https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/description/

Video Links

https://www.youtube.com/watch?v=MOe7LlagCN8&ab_channel=AryanMittal

Approach 1:

C++

class Solution {
public:
	int count_smaller(vector<vector<int>>& arr, int mid){
		int count = 0;
		for(int i=0; i<arr.size(); i++){
			count += upper_bound(arr[i].begin(), arr[i].end(), mid) - arr[i].begin();
		}

		return count;
	}

    int kthSmallest(vector<vector<int>>& arr, int k) {
		int m = arr.size(), n = arr[0].size();
		int low = arr[0][0], high = arr[m-1][n-1];
		int ans;
    
		while(low <= high){
			int mid = low + (high - low)/2;
			int ct = count_smaller(arr, mid);

			if(ct >= k){
				high = mid - 1;
				ans = mid;
			}
			else{
				low = mid + 1;
			}
		}

		return ans;
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++

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