Smallest Intersecting Element - Amazon Top Interview Questions


Problem Statement :


You are given a two-dimensional list of integers matrix where each row is sorted in ascending order. Return the smallest number that exists in every row. If there's no solution, return -1.

Constraints

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

Example 1

Input

matrix = [
    [1, 2, 4],
    [4, 9, 9],
    [0, 2, 4]
]

Output

4


Solution :



title-img 




                        Solution in C++ :

int solve(vector<vector<int>>& matrix) {
    int n = matrix.size();
    if (n == 0) return -1;
    int m = matrix[0].size();
    if (m == 0) return -1;
    vector<int> pointers(n);
    set<pair<int, int>> s;
    for (int i = 0; i < n; ++i) s.emplace(matrix[i][0], i);
    while (s.begin()->first < s.rbegin()->first) {
        int i = s.begin()->second;
        if (pointers[i] == m - 1) return -1;
        pointers[i]++;
        s.erase(s.begin());
        s.emplace(matrix[i][pointers[i]], i);
    }
    return s.begin()->first;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[][] matrix) {
        Map<Integer, Integer> map = new HashMap<>();

        for (int[] row : matrix) {
            int prev = Integer.MAX_VALUE;
            for (int v : row) {
                if (v == prev)
                    continue;
                map.put(v, map.getOrDefault(v, 0) + 1);
                prev = v;
            }
        }

        int res = Integer.MAX_VALUE;
        for (Map.Entry<Integer, Integer> en : map.entrySet()) {
            if (en.getValue().intValue() == matrix.length)
                res = Math.min(res, en.getKey().intValue());
        }
        return res == Integer.MAX_VALUE ? -1 : res;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def bi(self, li, val):
        i = 0
        j = len(li) - 1
        while i <= j:
            m = int((i + j) // 2)
            k = li[m]
            if k == val:
                return True
            elif k < val:
                i = m + 1
            else:
                j = m - 1
        return False

    def solve(self, matrix):
        for i in range(len(matrix)):
            matrix[i] = list(set(matrix[i]))
            matrix[i].sort()
        if len(matrix) == 0:
            return -1
        for i in matrix[0]:
            k = 1
            for j in matrix:

                if not self.bi(j, i):
                    # print(j,i)
                    k = 0
                    break
            if k == 1:
                return i
        return -1


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