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

• Difficulty Level : Easy
• Last Updated : 08 Apr, 2021

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++

 // 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 elementsint 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 Codeint main(){         // Given array    int a[] = { 1, 1, 1, 1, 1 };    int N = sizeof(a) / sizeof(a);     // Function call    cout << countElements(a, N);} // This code is contributed by Amit Katiyar

Java

 // 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

 # 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 elementsdef countElements(a, n):         # Stores the frequencies    # of array elements    cnt =  * (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 Codeif __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#

 // C# program for above approachusing System; class GFG{ // Function to find the number of array// elements that can be represented as the sum// of two or more consecutive array elementsstatic 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 Codepublic 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

Javascript


Output:
0

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

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