Given an array of n integers, count all different triplets whose sum is equal to the perfect cube i.e, for any i, j, k(i < j < k) satisfying the condition that a[i] + a[j] + a[j] = X^{3} where X is any integer. 3 ≤ n ≤ 1000, 1 ≤ a[i, j, k] ≤ 5000

Input: N = 5 2 5 1 20 6 Output: 3 Explanation: There are only 3 triplets whose total sum is a perfect cube. Indices Values SUM 0 1 2 2 5 1 8 0 1 3 2 5 20 27 2 3 4 1 20 6 27 Since 8 and 27 are prefect cube of 2 and 3.

**
Naive appraoch** is to iterate over all the possible numbers by using 3 nested loops and check whether their sum is perfect cube or not. The approach would be very slow as time complexity can go up to O(n

^{3}).

An **Efficient **approach is to use dynamic programming and basic mathematics. According to given condition sum of any of three positive integer is not greater than 15000. Therefore there can be only 24(15000^{1/3}) cubes are possible in the range of 1 to 15000.

Now instead of iterating all triplets we can do much better by the help of above information. Fixed first two indices i and j such that instead of iterating over all k(j < k ≤ n), we can iterate over all the 24 possible cubes, and for each one, (let's say P) check how many occurrence of **P – (a[i] + a[j])** are in a[j+1, j+2, … n].

But if we compute the number of occurance of a number say K in a[j+1, j+2, … n] then this would again be counted as slow approch and would definitely give TLE. So we have to think a different approach.

Now here comes to a Dynamic Programming. Since all numbers are smaller than 5000 and n is at most 1000. Hence we can compute a DP array as,

**dp[i][K]:= Number of occurance of K in A[i, i+1, i+2 … n]**

## C++

// C++ program to calculate all triplets whose // sum is perfect cube. #include <bits/stdc++.h> using namespace std; int dp[1001][15001]; // Function to calculate all occurrence of // a number in a given range void computeDpArray(int arr[], int n) { for (int i = 0; i < n; ++i) { for (int j = 1; j <= 15000; ++j) { // if i == 0 // assign 1 to present state if (i == 0) dp[i][j] = (j == arr[i]); // else add +1 to current state with // previous state else dp[i][j] = dp[i - 1][j] + (arr[i] == j); } } } // Function to calculate triplets whose sum // is equal to the pefect cube int countTripletSum(int arr[], int n) { computeDpArray(arr, n); int ans = 0; // Initialize answer for (int i = 0; i < n - 2; ++i) { for (int j = i + 1; j < n - 1; ++j) { for (int k = 1; k <= 24; ++k) { int cube = k * k * k; int rem = cube - (arr[i] + arr[j]); // count all occurrence of third triplet // in range from j+1 to n if (rem > 0) ans += dp[n - 1][rem] - dp[j][rem]; } } } return ans; } // Driver code int main() { int arr[] = { 2, 5, 1, 20, 6 }; int n = sizeof(arr) / sizeof(arr[0]); cout << countTripletSum(arr, n); return 0; }

## Java

// JAVA Code for Count all triplets whose // sum is equal to a perfect cube import java.util.*; class GFG { public static int dp[][]; // Function to calculate all occurrence of // a number in a given range public static void computeDpArray(int arr[], int n) { for (int i = 0; i < n; ++i) { for (int j = 1; j <= 15000; ++j) { // if i == 0 // assign 1 to present state if (i == 0 && j == arr[i]) dp[i][j] = 1; else if(i==0) dp[i][j] = 0; // else add +1 to current state // with previous state else if(arr[i] == j) dp[i][j] = dp[i - 1][j] + 1; else dp[i][j] = dp[i - 1][j]; } } } // Function to calculate triplets whose sum // is equal to the pefect cube public static int countTripletSum(int arr[], int n) { computeDpArray(arr, n); int ans = 0; // Initialize answer for (int i = 0; i < n - 2; ++i) { for (int j = i + 1; j < n - 1; ++j) { for (int k = 1; k <= 24; ++k) { int cube = k * k * k; int rem = cube - (arr[i] + arr[j]); // count all occurrence of // third triplet in range // from j+1 to n if (rem > 0) ans += dp[n - 1][rem] - dp[j][rem]; } } } return ans; } /* Driver program to test above function */ public static void main(String[] args) { int arr[] = { 2, 5, 1, 20, 6 }; int n = arr.length; dp = new int[1001][15001]; System.out.println(countTripletSum(arr, n)); } } // This code is contributed by Arnav Kr. Mandal.

**Output:**

3

**Time complexity: **O(N^{2}*24)

**Auxiliary space: **O(10^{7})

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