**Merging K Sorted Lists - Amazon Top Interview Questions**

### Problem Statement :

Given a list of sorted lists of integers, merge them into one large sorted list. Constraints 0 ≤ n * m ≤ 100,000 where n is number of rows and m is the longest column in lists Example 1 Input lists = [ [], [], [10, 12], [], [3, 3, 13], [3], [10], [0, 7] ] Output [0, 3, 3, 3, 7, 10, 10, 12, 13]

### Solution :

` ````
Solution in C++ :
vector<int> merge(vector<int>& a, vector<int>& b) {
vector<int> temp;
int i = 0, j = 0;
if (a.size() == 0) return b;
if (b.size() == 0) return a;
while (i < a.size() && j < b.size()) {
if (a[i] <= b[j]) {
temp.push_back(a[i]);
i++;
} else {
temp.push_back(b[j]);
j++;
}
}
while (i < a.size()) {
temp.push_back(a[i]);
i++;
}
while (j < b.size()) {
temp.push_back(b[j]);
j++;
}
return temp;
}
vector<int> helper(vector<vector<int>>& lists, int st, int end) {
if (st == end) return lists[st];
int mid = st + (end - st) / 2;
vector<int> a = helper(lists, st, mid);
vector<int> b = helper(lists, mid + 1, end);
return merge(a, b);
}
vector<int> solve(vector<vector<int>>& lists) {
int k = lists.size();
if (k == 0) return {};
return helper(lists, 0, k - 1);
}
```

` ````
Solution in Python :
class Solution:
def solve(self, lists):
min_heap = []
for i in range(len(lists)):
if len(lists[i]) > 0:
heapq.heappush(min_heap, (lists[i][0], i, 0))
res = []
while len(min_heap) > 0:
a, i, j = heapq.heappop(min_heap)
if j + 1 < len(lists[i]):
heapq.heappush(min_heap, (lists[i][j + 1], i, j + 1))
res.append(a)
return res
```

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