C++를 사용하여 최소값과 최대값이 동일한 하위 배열의 수를 찾는 코드를 작성하세요.

Input : array = { 2, 3, 6, 6, 2, 4, 4, 4 }
Output : 12
Explanation : {2}, {3}, {6}, {6}, {2}, {4}, {4}, {4}, {6,6}, {4,4}, {4,4} and { 4,4,4 } are the subarrays which can be formed with maximum and minimum element same.

Input : array = { 3,3,1,5,1,2,2 }
Output : 9
Explanation : {3}, {3}, {1}, {5}, {1}, {2}, {2}, {3,3} and {2,2} are the subarrays which can be formed with minimum and maximum are the same.

#include <bits/stdc++.h>
using namespace std;
int main(){
    int a[ ] = { 2, 4, 5, 3, 3, 3 };
    int n = sizeof(a) / sizeof(a[0]);
        int result = n, count =0;
    for (int i = 0; i < n; i++) {
        for (int j = i+1; j < n; j++) {
            if(a[i]==a[j])
                count++;
            else
                break;
        }
        result+=count;
        count =0;
    }
    cout << "Number of subarrays having minimum and maximum elements same:" << result;
    return 0;
}

Number of subarrays having minimum and maximum elements same: 9
Time complexity = O(n<sup>2</sup>).

#include <bits/stdc++.h>
using namespace std;
int main(){
    int a[] = { 2, 4, 5, 3, 3, 3 };
    int n = sizeof(a) / sizeof(a[0]);
        int result = 0;
    int count =1,temp=a[0];
    for (int i = 1; i < n; i++) {
        if (temp==a[i]){
            count++;
        }
        else{
            temp=a[i];
            result = result + (count*(count+1)/2);
            count=1;
        }
    }
    result = result + (count*(count+1)/2);
    cout <<  "Number of subarrays having minimum and maximum elements same:" << result;
    return 0;
}

Number of subarrays having minimum and maximum elements same: 9
Time complexity : O(n)