# List Consecutive Split - Google Top Interview Questions

### Problem Statement :

```Given a list of integers nums, and an integer k, return whether the list can be split into lists where each list contains k integers and is consecutively increasing.

Constraints

n ≤ 100,000 where n is the length of nums.

Example 1

Input

nums = [3, 2, 3, 4, 5, 1]

k = 3

Output

True

Explanation

We can split the list into [1, 2, 3] and [3, 4, 5]```

### Solution :

```                        ```Solution in C++ :

bool solve(vector<int>& nums, int k) {
map<int, int> m;
for (int n : nums) m[n]++;
while (m.size()) {
auto [x, y] = *m.begin();
if (y == 0) {
m.erase(x);
continue;
}
for (int i = x; i < x + k; i++) {
if (m[i] < y) {
return false;
}
m[i] -= y;
}
}
return m.size() == 0;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public boolean solve(int[] nums, int k) {
if (nums == null || nums.length == 0 || k > nums.length)
return false;
Map<Integer, Integer> freqMap = new HashMap();
for (int num : nums) freqMap.put(num, freqMap.getOrDefault(num, 0) + 1);
PriorityQueue<Integer> pq = new PriorityQueue<>();
for (int key : freqMap.keySet()) pq.offer(key);

while (!pq.isEmpty() && pq.size() >= k) {
int curr = -1;
List<Integer> temp = new ArrayList();
for (int i = 1; i <= k; i++) {
int num = pq.poll();
if (freqMap.get(num) == 1)
freqMap.remove(num);
else
freqMap.put(num, freqMap.get(num) - 1);
if (curr == -1) {
curr = num;
} else {
if (num != curr + 1)
return false;
curr = num;
}
}
for (int i = 0; i < temp.size(); i++) {
if (freqMap.containsKey(temp.get(i)))
pq.offer(temp.get(i));
}
}
return pq.isEmpty();
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, nums, k):
fre = Counter(nums)
# greedy solution, take the smallest then try
vals = list(fre.keys())
heapq.heapify(vals)

# every iteration i make a new list
while len(vals):
start = vals
# now create a list starting with 'start'
for i in range(k):
if fre[start + i]:
fre[start + i] -= 1
else:
return False
# pop any elements which are not remaining
while len(vals) and fre[vals] == 0:
heapq.heappop(vals)

return True```
```

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