# Program to check whether list can be split into sublists of k increasing elements in C++

C++Server Side ProgrammingProgramming

Suppose we have a list of numbers called nums, and another number k, we have to check whether the list can be split into lists where each list contains k values and the values are consecutively increasing.

So, if the input is like nums = [4, 3, 2, 4, 5, 6], k = 3, then the output will be True, as we can split the list into [2, 3, 4] and [4, 5, 6]

To solve this, we will follow these steps −

• Define one map

• for each key it in m

• increase m[it] by 1

• ok := true

• while (size of m is not 0 and ok is true), do −

• ok := false

• for each key-value pair it in m

• if (it.key - 1) is not in m, then −

• flag := true

• for initialize i := it.key, when i <= it.key + k - 1, update (increase i by 1), do −

• if i is not present in m, then −

• flag := false

• if flag is true, then −

• for initialize i := it.key , when i <= it.key + k - 1, update (increase i by 1), do −

• (decrease m[i] by 1)

• if m[i] is same as 0, then −

• delete i from m

• ok := true

• Come out from the loop

• return true when m is empty, otherwise false.

Let us see the following implementation to get better understanding −

## Example

Live Demo

#include<bits/stdc++.h>
using namespace std;
class Solution {
public:
bool solve(vector<int> nums, int k) {
map <int, int> m;
for(auto& it : nums){
m[it]++;
}
bool ok = true;
while(m.size() && ok){
ok = false;
for(auto& it : m){
if(!m.count(it.first - 1)){
bool flag = true;
for(int i = it.first; i <= it.first + k - 1;i++){
if(!m.count(i))
flag = false;
}
if(flag){
for(int i = it.first; i <= it.first + k - 1;i++){
m[i]--;
if(m[i] == 0)
m.erase(i);

}
ok = true;
break;
}
}
}
}
return m.empty();
}
};
main(){
vector<int> v = {4, 3, 2, 4, 5, 6};
Solution ob;
cout << ob.solve(v, 3);
}



## Input

{4, 3, 2, 4, 5, 6}

## Output

1