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Count Number of Nice Subarrays in C++
Suppose we have an array of integers nums and an integer k. A subarray is known as nice subarray if there are k odd numbers on it. We have to find the number of nice sub-arrays. So if the array is [1,1,2,1,1], and k = 3, then the output will be 2, as the subarrays are [1,1,2,1], and [1,2,1,1]
To solve this, we will follow these steps −
- ans := 0, n := size of nums array
- left := 0 and right := 0, and count := 0
- define an array odd, fill this with all odd values present in nums
- if length of odd array is >= k, then
- for i is 0 and j in range k – 1 to size of odd – 1, increase i and j by 1
- left := odd[i] + 1 if i = 0, otherwise odd[i] – odd[i – 1]
- right := odd[j] if size of odd – 1 = j, otherwise odd[j + 1] – odd[j]
- ans := ans + left * right
- for i is 0 and j in range k – 1 to size of odd – 1, increase i and j by 1
- return ans
Example(C++)
Let us see the following implementation to get a better understanding −
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int numberOfSubarrays(vector<int>& nums, int k) {
int ans = 0;
int n = nums.size();
int left = 0;
int right = 0;
int cnt = 0;
vector <int> odd;
for(int i = 0; i < n; i++){
if(nums[i] % 2 == 1)odd.push_back(i);
}
if(odd.size()>=k){
for(int i = 0, j = k-1; j < odd.size(); i++, j++){
int left = i==0?odd[i]+1: odd[i] - odd[i-1];
int right = j==odd.size()-1 ?n-odd[j] : odd[j+1] - odd[j];
ans += left * right;
}
}
return ans;
}
};
main(){
vector<int> v = {1,1,2,1,1};
Solution ob;
cout <<ob.numberOfSubarrays(v, 3);
}Input
[1,1,2,1,1] 3
Output
2
Updated on: 30-Apr-2020
730 Views
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