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Find duplicates under given constraints

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A sorted array contains 6 different numbers, only 1 number is repeated five times. So there are total 10 numbers in array. Find the duplicate numbers using two comparisons only. 

Examples : 

Input: arr[] = {1, 1, 1, 1, 1, 5, 7, 10, 20, 30}
Output: 1

Input: arr[] = {1, 2, 3, 3, 3, 3, 3, 5, 9, 10}
Output: 3

Asked in Yahoo

An important observation is, arr[4] or arr[5] is an occurrence of repeated element for sure. Below is the implementation based on this observation. 

Implementation:

CPP

// C++ program to find duplicate element under
// given constraints.
#include<bits/stdc++.h>
using namespace std;
 
// This function assumes array is sorted,  has
// 10 elements,  there are total 6 different
// elements and one element repeats 5 times.
int findDuplicate(int a[])
{
    if (a[3] == a[4])
        return a[3];
    else if (a[4] == a[5])
        return a[4];
    else
        return a[5];
}
 
// Driver code
int main()
{
    int a[] = {1, 1, 1, 1, 1, 5, 7, 10, 20, 30};
    cout << findDuplicate(a);
    return 0;
}

                    

Java

// Java program to find duplicate element under
// given constraints.
class Num{
 
// This function assumes array is sorted, has
// 10 elements, there are total 6 different
// elements and one element repeats 5 times.
static int findDuplicate(int a[])
{
    if (a[3] == a[4])
        return a[3];
    else if (a[4] == a[5])
        return a[4];
    else
        return a[5];
}
 
// Driver code
public static void main(String[] args)
{
    int a[] = {1, 1, 1, 1, 1, 5, 7, 10, 20, 30};
    System.out.println(findDuplicate(a));
}
}
//This code is contributed by
//Smitha Dinesh Semwal

                    

Python3

# Python 3 program to find duplicate
# element under given constraints.
 
# This function assumes array is sorted, has 10
# elements, there are total 6 different elements
# and one element repeats 5 times.
def findDuplicate(a):
 
    if (a[3] == a[4]):
        return a[3]
    elif (a[4] == a[5]):
        return a[4]
    else:
        return a[5]
 
# Driver code
a = [1, 1, 1, 1, 1, 5, 7, 10, 20, 30]
 
print(findDuplicate(a))
# This code is contributed by Yash Agarwal

                    

C#

// C# program to find duplicate 
// element under given constraints.
using System;
 
class GFG {
 
// This function assumes array is
// sorted, has 10 elements, there
// are total 6 different elements
// and one element repeats 5 times.
static int findDuplicate(int []a)
{
    if (a[3] == a[4])
        return a[3];
    else if (a[4] == a[5])
        return a[4];
    else
        return a[5];
}
 
// Driver code
public static void Main()
{
    int []a = {1, 1, 1, 1, 1, 5,
               7, 10, 20, 30};
    Console.Write(findDuplicate(a));
}
}
 
// This code is contributed by nitin mittal

                    

PHP

<?php
// PHP program to find duplicate
// element under given constraints.
 
// This function assumes array is
// sorted, has 10 elements, there
// are total 6 different elements
// and one element repeats 5 times.
function findDuplicate($a)
{
    if ($a[3] == $a[4])
        return $a[3];
    else if ($a[4] == $a[5])
        return $a[4];
    else
        return $a[5];
}
 
// Driver code
$a = array(1, 1, 1, 1, 1,
           5, 7, 10, 20, 30);
echo findDuplicate($a);
 
// This code is contributed by nitin mittal.
?>

                    

Javascript

<script>
 
// JavaScript program to find duplicate element under
// given constraints.
 
 
// This function assumes array is sorted, has
// 10 elements, there are total 6 different
// elements and one element repeats 5 times.
function findDuplicate(a)
{
    if (a[3] == a[4])
        return a[3];
    else if (a[4] == a[5])
        return a[4];
    else
        return a[5];
}
 
// Driver code
    let a = [1, 1, 1, 1, 1, 5, 7, 10, 20, 30];
    document.write(findDuplicate(a));
     
</script>

                    

Output
1

Time Complexity: O(1)

Auxiliary Space: O(1)

Exercise: Extend the above problem for an array with n different elements, size of array is 2*(n-1) and one element repeats (n-1) times.

 



Last Updated : 11 Feb, 2023
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