# 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.

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:

- 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])
- 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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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**.