# Count array elements that can be represented as sum of at least two consecutive array elements

Given an array A[] consisting of N integers from a range [1, N], the task is to calculate the count of array elements (non-distinct) that can be represented as the sum of two or more consecutive array elements.

Examples:

Input: a[] = {3, 1, 4, 1, 5, 9, 2, 6, 5}
Output: 5
Explanation:
The array elements satisfying the condition are:
4 = 3 + 1
5 = 1 + 4 or 4 + 1
9 = 3 + 1 + 4 + 1
6 = 1 + 5 or 1 + 4 + 1
5 = 1 + 4 or 4 + 1

Input: a[] = {1, 1, 1, 1, 1}
Output: 0
Explanation:
No such array element exists that can be represented as the sum of two or more consecutive elements.

Naive Approach: Traverse the given array for each element, find the sum of all possible subarrays and check if sum of any of the subarrays becomes equal to that of the current element. Increase count if found to be true. Finally, print the count obtained.

Time Complexity: O(n3
Auxiliary Space: O(1)

Efficient Approach: Follow the steps below to optimize the above approach:

• Initialize an array cnt[] to store the number of occurrences of each array element.
• Iterate over all subarrays of at least length 2 maintaining the sum of the current subarray sum.
• If the current sum does not exceed N, then add cnt[sum] to the answer and set cnt[sum]=0 to prevent counting the same elements several times.
• Finally, print the sum obtained.

Below is the implementation of the above approach:

 // C++ program for above approach  #include using namespace std;    // Function to find the number of // array elements that can be  // represented as the sum of two  // or more consecutive array elements int countElements(int a[], int n) {            // Stores the frequencies     // of array elements     int cnt[n + 1] = {0};     memset(cnt, 0, sizeof(cnt));            // Stores required count     int ans = 0;        // Update frequency of     // each array element     for(int i = 0; i < n; i++)     {         ++cnt[a[i]];     }            // Find sum of all subarrays     for(int l = 0; l < n; ++l)     {         int sum = 0;            for(int r = l; r < n; ++r)         {             sum += a[r];                if (l == r)                 continue;                if (sum <= n)             {                                    // Increment ans by cnt[sum]                 ans += cnt[sum];                    // Reset cnt[sum] by 0                 cnt[sum] = 0;             }         }     }        // Return ans     return ans; }    // Driver Code int main() {            // Given array     int a[] = { 1, 1, 1, 1, 1 };     int N = sizeof(a) / sizeof(a[0]);        // Function call     cout << countElements(a, N); }    // This code is contributed by Amit Katiyar

 // Java Program for above approach    import java.util.*; class GFG {        // Function to find the number of array     // elements that can be represented as the sum     // of two or more consecutive array elements     static int countElements(int[] a, int n)     {         // Stores the frequencies         // of array elements         int[] cnt = new int[n + 1];            // Stores required count         int ans = 0;            // Update frequency of         // each array element         for (int k : a) {             ++cnt[k];         }            // Find sum of all subarrays         for (int l = 0; l < n; ++l) {                int sum = 0;                for (int r = l; r < n; ++r) {                 sum += a[r];                    if (l == r)                     continue;                    if (sum <= n) {                        // Increment ans by cnt[sum]                     ans += cnt[sum];                        // Reset cnt[sum] by 0                     cnt[sum] = 0;                 }             }         }            // Return ans         return ans;     }        // Driver Code     public static void main(String[] args)     {         // Given array         int[] a = { 1, 1, 1, 1, 1 };            // Function call         System.out.println(             countElements(a, a.length));     } }

 # Python3 program for above approach    # Function to find the number of array # elements that can be represented as the sum # of two or more consecutive array elements def countElements(a, n):            # Stores the frequencies     # of array elements     cnt = [0] * (n + 1)        # Stores required count     ans = 0        # Update frequency of     # each array element     for k in a:         cnt[k] += 1        # Find sum of all subarrays     for l in range(n):         sum = 0            for r in range(l, n):             sum += a[r]                if (l == r):                 continue             if (sum <= n):                    # Increment ans by cnt[sum]                 ans += cnt[sum]                    # Reset cnt[sum] by 0                 cnt[sum] = 0        # Return ans     return ans    # Driver Code if __name__ == '__main__':        # Given array     a = [ 1, 1, 1, 1, 1 ]        # Function call     print(countElements(a, len(a)))    # This code is contributed by mohit kumar 29

 // C# program for above approach using System;    class GFG{    // Function to find the number of array // elements that can be represented as the sum // of two or more consecutive array elements static int countElements(int[] a, int n) {            // Stores the frequencies     // of array elements     int[] cnt = new int[n + 1];        // Stores required count     int ans = 0;        // Update frequency of     // each array element     foreach(int k in a)      {         ++cnt[k];     }        // Find sum of all subarrays     for(int l = 0; l < n; ++l)     {         int sum = 0;            for(int r = l; r < n; ++r)         {             sum += a[r];             if (l == r)                 continue;                if (sum <= n)             {                                    // Increment ans by cnt[sum]                 ans += cnt[sum];                    // Reset cnt[sum] by 0                 cnt[sum] = 0;             }         }     }        // Return ans     return ans; }    // Driver Code public static void Main(String[] args) {            // Given array     int[] a = { 1, 1, 1, 1, 1 };        // Function call     Console.WriteLine(countElements(a, a.Length)); } }    // This code is contributed by Amit Katiyar

Output:
0

Time Complexity: O(N2)
Auxiliary Space: O(N)

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