Given an array of n elements such that every element of the array is an integer in the range 1 to n, find the sum of all the distinct elements of the array.
Examples:
Input: arr[] = {5, 1, 2, 4, 6, 7, 3, 6, 7}
Output: 28
The distinct elements in the array are 1, 2,
3, 4, 5, 6, 7
Input: arr[] = {1, 1, 1}
Output: 1
The problem has appeared here as a general problem and the solution will work for the above case also. But a better approach is explained below.
The approach is to mark the occurrences of the array elements by making the elements at those indices negative. Example, a[0] = 1, a[1] = 1, a[2] = 1.
We check if a[abs(a[i])-1] is >=0, if yes, mark a[abs(a[i])-1] as negative. i.e. a[0] = 1 >=0, we mark a[1-1] as a[0] = -1. Next, a[1], check if (abs(a[1]-1) is +ve or not. If -ve, it means a[1] has already occurred before, else it is the first occurrence of this element.
Implementation:
C++
// C++ program to find sum of distinct elements #include <iostream> using namespace std; // Returns sum of distinct elements in arr[] assuming // that elements in a[] are in range from 1 to n. int sumOfDistinct(int a[], int n) { int sum = 0; for (int i = 0; i < n; i++) { // If element appears first time if (a[abs(a[i]) - 1] >= 0) { sum += abs(a[i]); a[abs(a[i]) - 1] *= -1; } } return sum; } // Driver code int main() { int a[] = { 5, 1, 2, 4, 6, 7, 3, 6, 7 }; int n = sizeof(a)/sizeof(a[0]); cout << sumOfDistinct(a, n) << endl; return 0; } |
Java
// JAVA program to find sum of distinct // elements in sorted order import java.io.*; import java.util.*; import java.math.*; class GFG{ // Returns sum of distinct elements in arr[] // assuming that elements in a[] are in // range from 1 to n. static int sumOfDistinct(int a[], int n) { int sum = 0; for (int i = 0; i < n; i++) { // If element appears first time if (a[(Math.abs(a[i]) - 1)] >= 0) { sum += Math.abs(a[i]); a[(Math.abs(a[i]) - 1)] *= -1; } } return sum; } // Driver code public static void main(String args[]) { int a[] = { 5, 1, 2, 4, 6, 7, 3, 6, 7 }; int n = a.length; System.out.println(sumOfDistinct(a, n) ); } } // This code is contributed by Nikita Tiwari. |
Python3
# Python program to find sum of distinct elements # in sorted order import math # Returns sum of distinct elements in arr[] # assuming that elements in a[] are in # range from 1 to n. def sumOfDistinct(a , n) : sum = 0 i = 0 while i < n: # If element appears first time if (a[abs(a[i]) - 1] >= 0) : sum = sum + abs(a[i]) a[abs(a[i]) - 1] = a[abs(a[i]) - 1] * (-1) i = i + 1 return sum; # Driver code a = [ 5, 1, 2, 4, 6, 7, 3, 6, 7 ] n = len(a) print (sumOfDistinct(a, n)) # This code is contributed by Nikita Tiwari. |
C#
// C# program to find sum of distinct // elements in sorted order using System; class GFG{ // Returns sum of distinct elements // in arr[] assuming that elements // in a[] are in range from 1 to n static int sumOfDistinct(int []a, int n) { int sum = 0; for (int i = 0; i < n; i++) { // If element appears first time if (a[(Math.Abs(a[i]) - 1)] >= 0) { sum += Math.Abs(a[i]); a[(Math.Abs(a[i]) - 1)] *= - 1; } } return sum; } // Driver code public static void Main() { int []a = {5, 1, 2, 4, 6, 7, 3, 6, 7}; int n = a.Length; Console.Write(sumOfDistinct(a, n)); } } // This code is contributed by Nitin Mittal |
PHP
<?php // PHP program to find sum of // distinct elements // Returns sum of distinct // elements in arr[] assuming // that elements in a[] are // in range from 1 to n. function sumOfDistinct($a, $n) { $sum = 0; for ($i = 0; $i < $n; $i++) { // If element appears first time if ($a[abs($a[$i]) - 1] >= 0) { $sum += abs($a[$i]); $a[abs($a[$i]) - 1] *= -1; } } return $sum; } // Driver code $a = array(5, 1, 2, 4, 6, 7, 3, 6, 7); $n = sizeof($a); echo sumOfDistinct($a, $n) ; // This code is contributed by nitin mittal ?> |
Javascript
<script> // java script program to find sum of // distinct elements // Returns sum of distinct // elements in arr[] assuming // that elements in a[] are // in range from 1 to n. function sumOfDistinct(a, n) { let sum = 0; for (let i = 0; i < n; i++) { // If element appears first time if (a[(Math.abs(a[i])) - 1] >= 0) { sum += Math.abs(a[i]); a[(Math.abs(a[i]) - 1)] *= -1; } } return sum; } // Driver code let a = [5, 1, 2, 4, 6, 7, 3, 6, 7]; let n = a.length; document.write( sumOfDistinct(a, n)); //contributed by bobby </script> |
28
Time complexity: O(n)
Auxiliary Space : O(1)
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