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 + 1Input: 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.
C++
// C++ program for above approach#include <bits/stdc++.h>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 arr[], int n){ int count = 0; for (int i = 0; i < n; i++) { int sum = arr[i]; bool flag = false; for (int j = 0; j < i; j++) { int reqSum = 0; for (int k = j; k < i; k++) { reqSum += arr[k]; if ((k - j + 1) >= 2 && reqSum == sum) { flag = true; count++; break; } } if (flag) break; } for (int j = i + 1; j < n && flag == false; j++) { int reqSum = 0; for (int k = j; k < n; k++) { reqSum += arr[k]; if ((k - j + 1) >= 2 && reqSum == sum) { flag = true; count++; break; } } if (flag) break; } } return count;}// Driver Codeint main(){ // Given array int a[] = { 3, 1, 4, 1, 5, 9, 2, 6, 5 }; int N = sizeof(a) / sizeof(a[0]); // Function call cout << countElements(a, N);} |
Java
/*package whatever //do not write package name here */import java.io.*;class GFG{ // Function to find the number of // array elements that can be // represented as the sum of two // or more consecutive array elements public static int countElements(int arr[], int n) { int count = 0; for (int i = 0; i < n; i++) { int sum = arr[i]; boolean flag = false; for (int j = 0; j < i; j++) { int reqSum = 0; for (int k = j; k < i; k++) { reqSum += arr[k]; if ((k - j + 1) >= 2 && reqSum == sum) { flag = true; count++; break; } } if (flag) break; } for (int j = i + 1; j < n && flag == false; j++) { int reqSum = 0; for (int k = j; k < n; k++) { reqSum += arr[k]; if ((k - j + 1) >= 2 && reqSum == sum) { flag = true; count++; break; } } if (flag) break; } } return count; } public static void main(String[] args) { // Given array int a[] = { 3, 1, 4, 1, 5, 9, 2, 6, 5 }; int N = a.length; // Function call System.out.println(countElements(a, N)); }}// This code is contributed by sourabhdalal0001. |
Python3
def countElements(arr, n): count = 0 for i in range(n): sum_ = arr[i] flag = False for j in range(i): reqSum = 0 for k in range(j, i): reqSum += arr[k] if (k - j + 1) >= 2 and reqSum == sum_: flag = True count += 1 break if flag: break for j in range(i + 1, n): reqSum = 0 for k in range(j, n): reqSum += arr[k] if (k - j + 1) >= 2 and reqSum == sum_: flag = True count += 1 break if flag: break return count# Given arrayarr = [3, 1, 4, 1, 5, 9, 2, 6, 5]n = len(arr)# Function callprint(countElements(arr, n)) |
C#
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 public static int countElements(int[] arr, int n) { int count = 0; for (int i = 0; i < n; i++) { int sum = arr[i]; bool flag = false; for (int j = 0; j < i; j++) { int reqSum = 0; for (int k = j; k < i; k++) { reqSum += arr[k]; if ((k - j + 1) >= 2 && reqSum == sum) { flag = true; count++; break; } } if (flag) break; } for (int j = i + 1; j < n && flag == false; j++) { int reqSum = 0; for (int k = j; k < n; k++) { reqSum += arr[k]; if ((k - j + 1) >= 2 && reqSum == sum) { flag = true; count++; break; } } if (flag) break; } } return count; } public static void Main(string[] args) { // Given array int[] a = { 3, 1, 4, 1, 5, 9, 2, 6, 5 }; int N = a.Length; // Function call Console.WriteLine(countElements(a, N)); }}// This code is contributed by divya_p123. |
Javascript
// Javascript 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 elementsfunction countElements(arr, n){ let count = 0; for (let i = 0; i < n; i++) { let sum = arr[i]; let flag = false; for (let j = 0; j < i; j++) { let reqSum = 0; for (let k = j; k < i; k++) { reqSum += arr[k]; if ((k - j + 1) >= 2 && reqSum == sum) { flag = true; count++; break; } } if (flag) break; } for (let j = i + 1; j < n && flag == false; j++) { let reqSum = 0; for (let k = j; k < n; k++) { reqSum += arr[k]; if ((k - j + 1) >= 2 && reqSum == sum) { flag = true; count++; break; } } if (flag) break; } } return count;}// Driver Code// Given arraylet a = [3, 1, 4, 1, 5, 9, 2, 6, 5 ];let N = a.length;// Function callconsole.log(countElements(a, N)); |
Output
5
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 <bits/stdc++.h>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[0]); // Function call cout << countElements(a, N);}// This code is contributed by Amit Katiyar |
Java
// Java Program for above approachimport 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 = [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 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
<script>// Javascript 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 elementsfunction countElements(a, n){ // Stores the frequencies // of array elements var cnt = Array(n + 1).fill(0); // Stores required count var ans = 0; // Update frequency of // each array element for(k = 0; k < n; k++) { cnt[a[k]]++; } // Find sum of all subarrays for(l = 0; l < n; ++l) { var sum = 0; for(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// Given arrayvar a = [ 1, 1, 1, 1, 1 ];// Function calldocument.write(countElements(a, a.length));// This code is contributed by todaysgaurav </script> |
Output:
0
Time Complexity: O(N2)
Auxiliary Space: O(N)
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