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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 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);` `    ``// 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 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#

 `// 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`

## Javascript

 ``
Output:
`0`

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

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