# Count all triplets whose sum is equal to a perfect cube

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] = X3 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.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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(n3).

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(150001/3) 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(N2*24)
Auxiliary space: O(107)

This article is contributed by Shubham Bansal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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