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Count of only repeated element in a sorted array of consecutive elements

  • Difficulty Level : Medium
  • Last Updated : 14 May, 2021
Geek Week

Given a sorted array of consecutive elements. The array has only one element repeated many times. The task is to find the length of the sequence of the repeated elements.
Expected Time Complexity: Less than 0(n)

Examples: 

Input  : arr[] = {1, 2, 3, 4, 4, 4, 5, 6}
Output : 4 3
Repeated element is 4 and it appears 3 times.

Input  : arr[] = {4, 4, 4, 4, 4}
Output : 4 5 

Input  : arr[] = {6, 7, 8, 9, 10, 10, 11}
Output : 10 2

Input  : arr[] = {6, 7, 8, 9, 10, 10, 10}
Output : 10 3 

We need to find two things:  

  1. The number of times the element repeats. If the array is sorted and if the max-difference of two adjacent elements is 1, then the length of the repeated sequence is n – 1 – (array[n-1] – array[0])
  2. Value of the element. To the value, we do Binary Search.
     

C++




// C++ program to find the only repeated element
// and number of times it appears
#include <bits/stdc++.h>
using namespace std;
 
// Assumptions : vector a is sorted, max-difference
// of two adjacent elements is 1
pair<int, int> sequence(const vector<int>& a)
{
    if (a.size() == 0)
        return {0, 0};
 
    int s = 0;
    int e = a.size() - 1;
    while (s < e)
    {
        int m = (s + e) / 2;
 
        // if a[m] = m + a[0], there is no
        // repeating character in [s..m]
        if (a[m] >= m + a[0])
            s = m + 1;
 
       // if a[m] < m + a[0], there is a
       // repeating character in [s..m]
        else
            e = m;
    }
    return {a[s], a.size() - (a[a.size() - 1] - a[0])};
}
 
// Driver code
int main()
{
    pair<int, int>p = sequence({ 1, 2, 3, 4, 4, 4, 5, 6 });
    cout << "Repeated element is " << p.first
         << ", it appears " << p.second << " times";
    return 0;
}

Java




// Java program to find the only repeated element
// and number of times it appears
 
import java.awt.Point;
import java.util.Arrays;
import java.util.Vector;
 
class Test
{
    // Assumptions : vector a is sorted, max-difference
    // of two adjacent elements is 1
    static Point sequence(Vector<Integer> a)
    {
        if (a.size() == 0)
            return new Point(0, 0);
      
        int s = 0;
        int e = a.size() - 1;
        while (s < e)
        {
            int m = (s + e) / 2;
      
            // if a[m] = m + a[0], there is no
            // repeating character in [s..m]
            if (a.get(m) >= m + a.get(0))
                s = m + 1;
      
           // if a[m] < m + a[0], there is a
           // repeating character in [s..m]
            else
                e = m;
        }
        return new Point(a.get(s), a.size() - (a.get(a.size() - 1) - a.get(0)));
    }
     
    // Driver method
    public static void main(String args[])
    {
        Integer array[] = new Integer[]{1, 2, 3, 4, 4, 4, 5, 6};
        Point p = sequence(new Vector<>(Arrays.asList(array)));
        System.out.println("Repeated element is " + p.x +
                           ", it appears " + p.y + " times");
    }
}

Python3




# Python3 program to find the
# only repeated element and
# number of times it appears
 
# Assumptions : vector a is sorted,
# max-difference of two adjacent
# elements is 1
def sequence(a):
    if (len(a) == 0):
        return [0, 0]
 
    s = 0
    e = len(a) - 1
    while (s < e):
        m = (s + e) // 2
 
        # if a[m] = m + a[0], there is no
        # repeating character in [s..m]
        if (a[m] >= m + a[0]):
            s = m + 1
 
        # if a[m] < m + a[0], there is a
        # repeating character in [s..m]
        else:
            e = m
    return [a[s], len(a) - (
                a[len(a) - 1] - a[0])]
 
# Driver code
p = sequence([1, 2, 3, 4, 4, 4, 5, 6])
print("Repeated element is", p[0],
             ", it appears", p[1], "times")
 
# This code is contributed by Mohit Kumar

C#




// C# program to find the only repeated element
// and number of times it appears
using System;
using System.Collections.Generic;
 
public class Point
{
    public int first, second;
 
        public Point(int first, int second)
        {
            this.first = first;
            this.second = second;
        }
     
     
}
public class Test
{
    // Assumptions : vector a is sorted, max-difference
    // of two adjacent elements is 1
    static Point sequence(List<int> a)
    {
        if (a.Count == 0)
            return new Point(0, 0);
     
        int s = 0;
        int e = a.Count - 1;
        while (s < e)
        {
            int m = (s + e) / 2;
     
            // if a[m] = m + a[0], there is no
            // repeating character in [s..m]
            if (a[m] >= m + a[0])
                s = m + 1;
     
        // if a[m] < m + a[0], there is a
        // repeating character in [s..m]
            else
                e = m;
        }
        return new Point(a[s], a.Count - (a[a.Count - 1] - a[0]));
    }
     
    // Driver code
    public static void Main(String []args)
    {
        int []array = new int[]{1, 2, 3, 4, 4, 4, 5, 6};
        Point p = sequence(new List<int>(array));
        Console.WriteLine("Repeated element is " + p.first +
                        ", it appears " + p.second + " times");
    }
}
 
// This code has been contributed by 29AjayKumar

Javascript




<script>
 
// Javascript program to find the only repeated element
// and number of times it appears
     
// Assumptions : vector a is sorted, max-difference
    // of two adjacent elements is 1   
function sequence(a)
{
    if (a.length == 0)
        return [0, 0]
   
    let s = 0
    let e = a.length - 1
    while (s < e)
    {
        let m = Math.floor((s + e) / 2);
   
        // if a[m] = m + a[0], there is no
        // repeating character in [s..m]
        if (a[m] >= m + a[0])
            s = m + 1
   
        // if a[m] < m + a[0], there is a
        // repeating character in [s..m]
        else
            e = m
    }
    return [a[s], a.length - (
                a[a.length - 1] - a[0])]
}
 
// Driver method
let p = sequence([1, 2, 3, 4, 4, 4, 5, 6])
document.write("Repeated element is "+ p[0]+
             ", it appears "+ p[1]+ " times")
     
 
// This code is contributed by patel2127
</script>

Output:  

Repeated element is 4, it appears 3 times

This article is contributed by Rakesh Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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