Sum of distinct elements when elements are in range 1 to n

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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++ 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 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.

# 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# 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 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
?>

<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)