Given a sorted array of consecutive elements. The array has only one element repeated many times. The task is to find length of the sequence of repeated element.

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:

- Number of times the element repeats. If the array is sorted and if 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; ` `} ` |

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## 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"` `); ` ` ` `} ` `} ` |

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## 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 ` |

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## 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 ` |

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**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 contribute.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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