Given a number n, find the number of ways in which we can add 3 non-negative integers so that their sum is n.
Examples :
Input : n = 1 Output : 3 There are three ways to get sum 1. (1, 0, 0), (0, 1, 0) and (0, 0, 1) Input : n = 2 Output : 6 There are six ways to get sum 2. (2, 0, 0), (0, 2, 0), (0, 0, 2), (1, 1, 0) (1, 0, 1) and (0, 1, 1) Input : n = 3 Output : 10 There are ten ways to get sum 3. (3, 0, 0), (0, 3, 0), (0, 0, 3), (1, 2, 0) (1, 0, 2), (0, 1, 2), (2, 1, 0), (2, 0, 1), (0, 2, 1) and (1, 1, 1)
Method 1 [ Brute Force : O(n3) ]
A simple solution is to consider all triplets using three loops. For every triplet, check if its sum is equal to n. If the sum is n, increment the count of solutions.
Below is the implementation.
C++
// A naive C++ solution to count solutions of// a + b + c = n#include<bits/stdc++.h>using namespace std;// Returns count of solutions of a + b + c = nint countIntegralSolutions(int n){ // Initialize result int result = 0; // Consider all triplets and increment // result whenever sum of a triplet is n. for (int i=0; i<=n; i++) for (int j=0; j<=n-i; j++) for (int k=0; k<=(n-i-j); k++) if (i + j + k == n) result++; return result;}// Driver codeint main(){ int n = 3; cout << countIntegralSolutions(n); return 0;} |
Java
// A naive Java solution to count// solutions of a + b + c = nimport java.io.*;class GFG { // Returns count of solutions of a + b + c = n static int countIntegralSolutions(int n) { // Initialize result int result = 0; // Consider all triplets and increment // result whenever sum of a triplet is n. for (int i = 0; i <= n; i++) for (int j = 0; j <= n - i; j++) for (int k = 0; k <= (n - i - j); k++) if (i + j + k == n) result++; return result; } // Driver code public static void main (String[] args) { int n = 3; System.out.println( countIntegralSolutions(n)); }}// This article is contributed by vt_m |
Python3
# Python3 code to count # solutions of a + b + c = n# Returns count of solutions# of a + b + c = ndef countIntegralSolutions (n): # Initialize result result = 0 # Consider all triplets and # increment result whenever # sum of a triplet is n. for i in range(n + 1): for j in range(n + 1): for k in range(n + 1): if i + j + k == n: result += 1 return result # Driver coden = 3print(countIntegralSolutions(n))# This code is contributed by "Sharad_Bhardwaj". |
C#
// A naive C# solution to count// solutions of a + b + c = nusing System;class GFG { // Returns count of solutions // of a + b + c = n static int countIntegralSolutions(int n) { // Initialize result int result = 0; // Consider all triplets and increment // result whenever sum of a triplet is n. for (int i = 0; i <= n; i++) for (int j = 0; j <= n - i; j++) for (int k = 0; k <= (n - i - j); k++) if (i + j + k == n) result++; return result; } // Driver code public static void Main () { int n = 3; Console.Write(countIntegralSolutions(n)); }}// This article is contributed by Smitha. |
PHP
<?php// A naive PHP solution// to count solutions of// a + b + c = n// Returns count of // solutions of a + b + c = nfunction countIntegralSolutions( $n){ // Initialize result $result = 0; // Consider all triplets and increment // result whenever sum of a triplet is n. for ($i = 0; $i <= $n; $i++) for ($j = 0; $j <= $n - $i; $j++) for ($k = 0; $k <= ($n - $i - $j); $k++) if ($i + $j + $k == $n) $result++; return $result;} // Driver Code $n = 3; echo countIntegralSolutions($n);// This code is contributed by anuj_67.?> |
Javascript
<script>// Javascript program solution to count// solutions of a + b + c = n // Returns count of solutions of a + b + c = n function countIntegralSolutions(n) { // Initialize result let result = 0; // Consider all triplets and increment // result whenever sum of a triplet is n. for (let i = 0; i <= n; i++) for (let j = 0; j <= n - i; j++) for (let k = 0; k <= (n - i - j); k++) if (i + j + k == n) result++; return result; }// Driver Code let n = 3; document.write(countIntegralSolutions(n));</script> |
Output:
10
Time complexity: O(n3)
Auxiliary Space: O(1)
Method 2 [ Direct Formula : O(1) ]
If we take a closer look at the pattern, we can find that the count of solutions is ((n+1) * (n+2)) / 2. The problem is equivalent to distributing n identical balls in three boxes and the solution is n+2C2. In general, if there are m variables (or boxes) and n balls , the formula becomes n+m-1Cm-1.
C++
// A naive C++ solution to count solutions of// a + b + c = n#include<bits/stdc++.h>using namespace std;// Returns count of solutions of a + b + c = nint countIntegralSolutions(int n){ return ((n+1)*(n+2))/2;}// Driver codeint main(){ int n = 3; cout << countIntegralSolutions(n); return 0;} |
Java
// Java solution to count // solutions of a + b + c = nimport java.io.*;class GFG { // Returns count of solutions // of a + b + c = n static int countIntegralSolutions(int n) { return ((n + 1) * (n + 2)) / 2; } // Driver code public static void main (String[] args) { int n = 3; System.out.println ( countIntegralSolutions(n)); }}// This article is contributed by vt_m |
Python3
# Python3 solution to count # solutions of a + b + c = n# Returns count of solutions# of a + b + c = ndef countIntegralSolutions (n): return int(((n + 1) * (n + 2)) / 2) # Driver coden = 3print(countIntegralSolutions(n))# This code is contributed by "Sharad_Bhardwaj". |
C#
// C# solution to count // solutions of a + b + c = nusing System;class GFG { // Returns count of solutions // of a + b + c = n static int countIntegralSolutions(int n) { return ((n + 1) * (n + 2)) / 2; } // Driver code public static void Main (String[] args) { int n = 3; Console.Write ( countIntegralSolutions(n)); }}// This code is contributed by parashar. |
PHP
<?php// A naive PHP solution // to count solutions of// a + b + c = n// Returns count of solutions// of a + b + c = nfunction countIntegralSolutions($n){ return (($n + 1) * ($n + 2)) / 2;} // Driver Code $n = 3; echo countIntegralSolutions($n);// This code is contributed by anuj_67.?> |
Javascript
<script>// A naive JavaScript solution // to count solutions of// a + b + c = n// Returns count of solutions of a + b + c = nfunction countIntegralSolutions(n){ return Math.floor(((n+1)*(n+2))/2);}// Driver code let n = 3; document.write(countIntegralSolutions(n));// This code is contributed by Surbhi Tyagi.</script> |
Output :
10
Time complexity: O(1)
Auxiliary Space: O(1)
This article is contributed by Shivam Gupta. If you like neveropen and would like to contribute, you can also write an article and mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
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