**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 :

` ````
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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