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.


Naive approach
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 occurrence of K in A[i, i+1, i+2 … n]

C++



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// 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;
}

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Java

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// 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.    

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Python3

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# 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

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C#

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// 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

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PHP

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<?php 
// PHP program to calculate all triplets 
// whose sum is perfect cube.
  
$dp = array_fill(0, 1001, 
      array_fill(0, 15001, NULL));
  
// Function to calculate all occurrence 
// of a number in a given range
function computeDpArray(&$arr, $n)
{
    global $dp;
    for ($i = 0; $i < $n; ++$i)
    {
        for ($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
function countTripletSum(&$arr, $n)
{
    global $dp;
    computeDpArray($arr, $n);
      
    $ans = 0; // Initialize answer
    for ($i = 0; $i < $n - 2; ++$i
    {
        for ($j = $i + 1; $j < $n - 1; ++$j
        {
            for ($k = 1; $k <= 24; ++$k
            {
                $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
$arr = array(2, 5, 1, 20, 6);
$n = sizeof($arr);
echo countTripletSum($arr, $n);
  
// This code is contributed by ita_c
?>

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Output:

 3

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

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