Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

N/3 repeated number in an array with O(1) space

  • Difficulty Level : Hard
  • Last Updated : 27 May, 2021

We are given a read only array of n integers. Find any element that appears more than n/3 times in the array in linear time and constant additional space. If no such element exists, return -1.

Examples:  

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Input : [10, 10, 20, 30, 10, 10]
Output : 10
10 occurs 4 times which is more than 6/3.

Input : [20, 30, 10, 10, 5, 4, 20, 1, 2]
Output : -1

The idea is based on Moore’s Voting algorithm.  We first find two candidates. Then we check if any of these two candidates is actually a majority. Below is the solution for above approach. 
 

C++




// CPP program to find if any element appears
// more than n/3.
#include <bits/stdc++.h>
using namespace std;
 
int appearsNBy3(int arr[], int n)
{
    int count1 = 0, count2 = 0;
    int first=INT_MAX    , second=INT_MAX    ;
 
    for (int i = 0; i < n; i++) {
 
        // if this element is previously seen,
        // increment count1.
        if (first == arr[i])
            count1++;
 
        // if this element is previously seen,
        // increment count2.
        else if (second == arr[i])
            count2++;
     
        else if (count1 == 0) {
            count1++;
            first = arr[i];
        }
 
        else if (count2 == 0) {
            count2++;
            second = arr[i];
        }
 
        // if current element is different from
        // both the previously seen variables,
        // decrement both the counts.
        else {
            count1--;
            count2--;
        }
    }
 
    count1 = 0;
    count2 = 0;
 
    // Again traverse the array and find the
    // actual counts.
    for (int i = 0; i < n; i++) {
        if (arr[i] == first)
            count1++;
 
        else if (arr[i] == second)
            count2++;
    }
 
    if (count1 > n / 3)
        return first;
 
    if (count2 > n / 3)
        return second;
 
    return -1;
}
 
int main()
{
    int arr[] = { 1, 2, 3, 1, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << appearsNBy3(arr, n) << endl;
    return 0;
}

Java




// Java program to find if any element appears
// more than n/3.
class GFG {
     
    static int appearsNBy3(int arr[], int n)
    {
        int count1 = 0, count2 = 0;
         
        // take the integers as the maximum
        // value of integer hoping the integer
        // would not be present in the array
        int first =  Integer.MIN_VALUE;;
        int second = Integer.MAX_VALUE;
     
        for (int i = 0; i < n; i++) {
     
            // if this element is previously
            // seen, increment count1.
            if (first == arr[i])
                count1++;
     
            // if this element is previously
            // seen, increment count2.
            else if (second == arr[i])
                count2++;
         
            else if (count1 == 0) {
                count1++;
                first = arr[i];
            }
     
            else if (count2 == 0) {
                count2++;
                second = arr[i];
            }
     
            // if current element is different
            // from both the previously seen
            // variables, decrement both the
            // counts.
            else {
                count1--;
                count2--;
            }
        }
     
        count1 = 0;
        count2 = 0;
     
        // Again traverse the array and
        // find the actual counts.
        for (int i = 0; i < n; i++) {
            if (arr[i] == first)
                count1++;
     
            else if (arr[i] == second)
                count2++;
        }
     
        if (count1 > n / 3)
            return first;
     
        if (count2 > n / 3)
            return second;
     
        return -1;
    }
     
    // Driver code
    public static void main(String args[])
    {
        int arr[] = { 1, 2, 3, 1, 1 };
        int n = arr.length;
        System.out.println(appearsNBy3(arr, n));
         
    }
}
 
// This code is contributed by Arnab Kundu

Python 3




# Python 3 program to find if
# any element appears more than
# n/3.
import sys
 
def appearsNBy3(arr, n):
 
    count1 = 0
    count2 = 0
    first = sys.maxsize
    second = sys.maxsize
 
    for i in range(0, n):
 
        # if this element is
        # previously seen,
        # increment count1.
        if (first == arr[i]):
            count1 += 1
 
        # if this element is
        # previously seen,
        # increment count2.
        elif (second == arr[i]):
            count2 += 1
     
        elif (count1 == 0):
            count1 += 1
            first = arr[i]
 
        elif (count2 == 0):
            count2 += 1
            second = arr[i]
         
 
        # if current element is
        # different from both
        # the previously seen
        # variables, decrement
        # both the counts.
        else:
            count1 -= 1
            count2 -= 1
         
     
 
    count1 = 0
    count2 = 0
 
    # Again traverse the array
    # and find the actual counts.
    for i in range(0, n):
        if (arr[i] == first):
            count1 += 1
 
        elif (arr[i] == second):
            count2 += 1
     
 
    if (count1 > n / 3):
        return first
 
    if (count2 > n / 3):
        return second
 
    return -1
 
# Driver code
arr = [1, 2, 3, 1, 1 ]
n = len(arr)
print(appearsNBy3(arr, n))
 
# This code is contributed by
# Smitha

C#




// C# program to find if any element appears
// more than n/3.
using System;
 
public class GFG {
     
    static int appearsNBy3(int []arr, int n)
    {
        int count1 = 0, count2 = 0;
         
        // take the integers as the maximum
        // value of integer hoping the integer
        // would not be present in the array
        int first = int.MaxValue;
        int second = int.MaxValue;
     
        for (int i = 1; i < n; i++) {
     
            // if this element is previously
            // seen, increment count1.
            if (first == arr[i])
                count1++;
     
            // if this element is previously
            // seen, increment count2.
            else if (second == arr[i])
                count2++;
         
            else if (count1 == 0) {
                count1++;
                first = arr[i];
            }
     
            else if (count2 == 0) {
                count2++;
                second = arr[i];
            }
     
            // if current element is different
            // from both the previously seen
            // variables, decrement both the
            // counts.
            else {
                count1--;
                count2--;
            }
        }
     
        count1 = 0;
        count2 = 0;
     
        // Again traverse the array and
        // find the actual counts.
        for (int i = 0; i < n; i++) {
            if (arr[i] == first)
                count1++;
     
            else if (arr[i] == second)
                count2++;
        }
     
        if (count1 > n / 3)
            return first;
     
        if (count2 > n / 3)
            return second;
     
        return -1;
    }
     
    // Driver code
    static public void Main(String []args)
    {
        int []arr = { 1, 2, 3, 1, 1 };
        int n = arr.Length;
        Console.WriteLine(appearsNBy3(arr, n));
    }
}
 
// This code is contributed by Arnab Kundu

PHP




<?php
// PHP program to find if any element appears
// more than n/3.
 
function appearsNBy3( $arr$n)
{
     $count1 = 0; $count2 = 0;
    $first = PHP_INT_MAX ; $second = PHP_INT_MAX ;
 
    for ( $i = 0; $i < $n; $i++) {
 
        // if this element is previously seen,
        // increment count1.
        if ($first == $arr[$i])
            $count1++;
 
        // if this element is previously seen,
        // increment count2.
        else if ($second == $arr[$i])
            $count2++;
     
        else if ($count1 == 0) {
            $count1++;
            $first = $arr[$i];
        }
 
        else if ($count2 == 0) {
            $count2++;
            $second = $arr[$i];
        }
 
        // if current element is different from
        // both the previously seen variables,
        // decrement both the counts.
        else {
            $count1--;
            $count2--;
        }
    }
 
    $count1 = 0;
    $count2 = 0;
 
    // Again traverse the array and find the
    // actual counts.
    for ($i = 0; $i < $n; $i++) {
        if ($arr[$i] == $first)
            $count1++;
 
        else if ($arr[$i] == $second)
            $count2++;
    }
 
    if ($count1 > $n / 3)
        return $first;
 
    if ($count2 > $n / 3)
        return $second;
 
    return -1;
}
 
// Driver code
$arr = array( 1, 2, 3, 1, 1 );
$n = count($arr);
echo appearsNBy3($arr, $n) ;
 
// This code is contributed by anuj_67.
?>

Javascript




<script>
    // Javascript program to find if any element appears more than n/3.
     
    function appearsNBy3(arr, n)
    {
        let count1 = 0, count2 = 0;
          
        // take the integers as the maximum
        // value of integer hoping the integer
        // would not be present in the array
        let first = Number.MAX_VALUE;
        let second = Number.MAX_VALUE;
      
        for (let i = 1; i < n; i++) {
      
            // if this element is previously
            // seen, increment count1.
            if (first == arr[i])
                count1++;
      
            // if this element is previously
            // seen, increment count2.
            else if (second == arr[i])
                count2++;
          
            else if (count1 == 0) {
                count1++;
                first = arr[i];
            }
      
            else if (count2 == 0) {
                count2++;
                second = arr[i];
            }
      
            // if current element is different
            // from both the previously seen
            // variables, decrement both the
            // counts.
            else {
                count1--;
                count2--;
            }
        }
      
        count1 = 0;
        count2 = 0;
      
        // Again traverse the array and
        // find the actual counts.
        for (let i = 0; i < n; i++) {
            if (arr[i] == first)
                count1++;
      
            else if (arr[i] == second)
                count2++;
        }
      
        if (count1 > parseInt(n / 3, 10))
            return first;
      
        if (count2 > parseInt(n / 3, 10))
            return second;
      
        return -1;
    }
     
    let arr = [ 1, 2, 3, 1, 1 ];
    let n = arr.length;
    document.write(appearsNBy3(arr, n));
 
// This cde is contributed by divyeshrabadiya07.
</script>
Output: 
1

 

Complexity Analysis:

  • Time Complexity:  O(n)
    First pass of the algorithm takes complete traversal over the array contributing to O(n) and another O(n) is consumed in checking if count1 and count2 is greater than floor(n/3) times.
  • Space Complexity: O(1)
    As no extra space is required so space complexity is constant



My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!