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

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

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 

Count of only repeated elements in a sorted array of consecutive elements using binary search:

To solve the problem follow the below idea:

We need to find two things:

• To find the number of times the element repeats: 
  • Suppose there is no repeated element for given array, then the least element will be at arr[0] and highest element will be at arr[n-1].
  • Now each time an element is repeated, the highest element will decrease by 1 each time.
  • Based on this idea, since the array is sorted and max-difference of two adjacent elements is 1, then: 

count of unique elements = arr[n-1] – arr[0]
Therefore, length of the repeated element = n – count of unique elements 
                                                                      = n – (array[n-1] – array[0])

• To find value of repeated element:
  • To find the repeated value, we can use Binary Search

Below is the implementation of the above approach:

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 });

    // Function call
    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 code
    public static void main(String args[])
    {
        Integer array[]
            = new Integer[] { 1, 2, 3, 4, 4, 4, 5, 6 };

        // Function call
        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
if __name__ == "__main__":
    p = sequence([1, 2, 3, 4, 4, 4, 5, 6])

    # Function call
    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));

        // Function call
        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

Time Complexity: O(log N)
Auxiliary Space: O(1)

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