Matrix Search - Amazon Top Interview Questions


Problem Statement :


Given a two-dimensional integer matrix, where every row and column is sorted in ascending order, find the kth (0-indexed) smallest number.

Constraints

n, m ≤ 250 where n and m are the number of rows and columns in matrix


Example 1


Input
matrix = [
    [1, 3, 30],
    [2, 3, 31],
    [5, 5, 32]
]
k = 4


Output
5




Example 2

Input

matrix = [
    [1, 2, 3]
]
k = 0


Output
1




Example 3


Input
matrix = [
    [1],
    [2],
    [3]
]
k = 2



Output
3


Solution :



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                        Solution in C++ :

int solve(vector<vector<int>>& A, int k) {
    int M = A.size();
    int N = A[0].size();

    int t = N * M - k;

    int l = 0;
    int h = 1e7;

    while (l <= h) {
        int m = (l + h) >> 1;

        int cnt = 0;

        for (auto x : A) cnt += x.end() - upper_bound(x.begin(), x.end(), m);

        if (cnt < t)
            h = m - 1;
        else
            l = m + 1;
    }

    return l;
}
                    

                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[][] mat, int k) {
        int left = mat[0][0];
        int right = mat[mat.length - 1][mat[0].length - 1];
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (possible(mat, k, mid)) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
    private boolean possible(int[][] mat, int k, int ele) {
        int i = 0;
        int j = mat[0].length - 1;
        int count = 0;
        while (i < mat.length && j >= 0) {
            if (mat[i][j] > ele)
                j--;
            else {
                count += (j + 1);
                i++;
            }
        }
        return count > k;
    }
}
                    

                        Solution in Python : 
                            
class Solution:
    def solve(self, matrix, n):
        def possible(x):
            # Is the number of elements y with
            # y <= x greater than n?
            c = len(matrix[0]) - 1
            count = 0
            for row in matrix:
                while c >= 0 and row[c] > x:
                    c -= 1
                count += c + 1
            return count > n

        lo = matrix[0][0]
        hi = matrix[-1][-1]
        while lo < hi:
            mi = lo + hi >> 1
            if not possible(mi):
                lo = mi + 1
            else:
                hi = mi
        return lo
                    

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