# 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
Example:

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 a 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 the given condition sum of any of three positive integers 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 occurrences of a number say K in a[j+1, j+2, … n] then this would again be counted as slow approach and would definitely give TLE. So we have to think of 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 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.

## Python3

 # Python 3 program to calculate all  # triplets whose sum is perfect cube.    dp = [[0 for i in range(15001)]           for j in range(1001)]    # Function to calculate all occurrence  # of a number in a given range def computeDpArray(arr, n):     for i in range(n):         for j in range(1, 15001, 1):                            # 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 def countTripletSum(arr, n):     computeDpArray(arr, n)            ans = 0 # Initialize answer     for i in range(0, n - 2, 1):         for j in range(i + 1, n - 1, 1):             for k in range(1, 25, 1):                 cube = k * k * k                    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 if __name__ == '__main__':     arr = [2, 5, 1, 20, 6]     n = len(arr)     print(countTripletSum(arr, n))    # This code is contributed by # Sahil_Shelangia

## C#

 // C# Code for Count all triplets whose // sum is equal to a perfect cube using System;    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 Code public static void Main()  {     int []arr = { 2, 5, 1, 20, 6 };     int n = arr.Length;     dp = new int[1001, 15001];            Console.Write(countTripletSum(arr, n)); } }    // This code is contributed // by 29AjayKumar

## PHP

 0)                     \$ans += \$dp[\$n - 1][\$rem] -                              \$dp[\$j][\$rem];             }         }     }     return \$ans; }    // Driver code \$arr = array(2, 5, 1, 20, 6); \$n = sizeof(\$arr); echo countTripletSum(\$arr, \$n);    // This code is contributed by ita_c ?>

Output:

3

Time complexity: O(N2*24)
Auxiliary space: O(107)