# Smallest subarray having an element with frequency greater than that of other elements

Given an array arr of positive integers, the task is to find the smallest length subarray of length more than 1 having an element occurring more times than any other element.

Examples:

Input: arr[] = {2, 3, 2, 4, 5}
Output: 2 3 2
Explanation: The subarray {2, 3, 2} has an element 2 which occurs more number of times any other element in the subarray.

Input: arr[] = {2, 3, 4, 5, 2, 6, 7, 6}
Output: 6 7 6
Explanation: The subarrays {2, 3, 4, 5, 2} and {6, 7, 6} contain an element that occurs more number of times than any other element in them. But the subarray {6, 7, 6} is of minimum length.

Naive Approach: A naive approach to solve the problem can be to find all the subarrays which have an element that meets the given condition and then finding the minimum of all those subarrays.

Time Complexity: O(N2
Auxiliary Space: O(N)

Efficient Approach: The problem can be reduced to find out that if there is any element occurring twice in a subarray, then it can be a potential answer. Because the minimum length of such subarray can be the minimum distance between two same elements

The idea is to use an extra array that maintains the last occurrence of the elements in the given array. Then find the distance between the last occurrence of an element and current position and find the minimum of all such lengths.

Below is the implementation of the above approach.

## C++

 `// C++ program for the above approach ` `#include   ` `using` `namespace` `std; ` ` `  `// Function to find subarray ` `void` `FindSubarray(``int` `arr[], ``int` `n) ` `{ ` `    ``// If the array has only one element, ` `    ``// then there is no answer. ` `    ``if` `(n == 1) { ` `        ``cout << ``"No such subarray!"` `             ``<< endl; ` `    ``} ` ` `  `    ``// Array to store the last occurrences ` `    ``// of the elements of the array. ` `    ``int` `vis[n + 1]; ` `    ``memset``(vis, -1, ``sizeof``(vis)); ` `    ``vis[arr] = 0; ` ` `  `    ``// To maintain the length ` `    ``int` `len = INT_MAX, flag = 0; ` ` `  `    ``// Variables to store ` `    ``// start and end indices ` `    ``int` `start, end; ` ` `  `    ``for` `(``int` `i = 1; i < n; i++) { ` `        ``int` `t = arr[i]; ` ` `  `        ``// Check if element is occurring ` `        ``// for the second time in the array ` `        ``if` `(vis[t] != -1) { ` `            ``// Find distance between last ` `            ``// and current index ` `            ``// of the element. ` `            ``int` `distance = i - vis[t] + 1; ` ` `  `            ``// If the current distance ` `            ``// is less then len ` `            ``// update len and ` `            ``// set 'start' and 'end' ` `            ``if` `(distance < len) { ` `                ``len = distance; ` `                ``start = vis[t]; ` `                ``end = i; ` `            ``} ` `            ``flag = 1; ` `        ``} ` ` `  `        ``// Set the last occurrence ` `        ``// of current element to be 'i'. ` `        ``vis[t] = i; ` `    ``} ` ` `  `    ``// If flag is equal to 0, ` `    ``// it means there is no answer. ` `    ``if` `(flag == 0) ` `        ``cout << ``"No such subarray!"` `             ``<< endl; ` `    ``else` `{ ` `        ``for` `(``int` `i = start; i <= end; i++) ` `            ``cout << arr[i] << ``" "``; ` `    ``} ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 2, 3, 2, 4, 5 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``FindSubarray(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program for the above approach ` `import` `java.util.*; ` ` `  `class` `GFG{ ` `     `  `// Function to find subarray  ` `public` `static` `void` `FindSubarray(``int``[] arr,  ` `                                ``int` `n)  ` `{  ` `     `  `    ``// If the array has only one element,  ` `    ``// then there is no answer.  ` `    ``if` `(n == ``1``) ` `    ``{  ` `        ``System.out.println(``"No such subarray!"``); ` `    ``}  ` ` `  `    ``// Array to store the last occurrences  ` `    ``// of the elements of the array.  ` `    ``int``[] vis = ``new` `int``[n + ``1``];  ` `    ``Arrays.fill(vis, -``1``); ` `    ``vis[arr[``0``]] = ``0``;  ` ` `  `    ``// To maintain the length  ` `    ``int` `len = Integer.MAX_VALUE, flag = ``0``;  ` ` `  `    ``// Variables to store  ` `    ``// start and end indices  ` `    ``int` `start = ``0``, end = ``0``;  ` ` `  `    ``for``(``int` `i = ``1``; i < n; i++) ` `    ``{  ` `        ``int` `t = arr[i];  ` ` `  `        ``// Check if element is occurring  ` `        ``// for the second time in the array  ` `        ``if` `(vis[t] != -``1``) ` `        ``{ ` `             `  `            ``// Find distance between last  ` `            ``// and current index  ` `            ``// of the element.  ` `            ``int` `distance = i - vis[t] + ``1``;  ` ` `  `            ``// If the current distance  ` `            ``// is less then len  ` `            ``// update len and  ` `            ``// set 'start' and 'end'  ` `            ``if` `(distance < len) ` `            ``{  ` `                ``len = distance;  ` `                ``start = vis[t];  ` `                ``end = i;  ` `            ``}  ` `            ``flag = ``1``;  ` `        ``}  ` ` `  `        ``// Set the last occurrence  ` `        ``// of current element to be 'i'.  ` `        ``vis[t] = i;  ` `    ``}  ` `     `  `    ``// If flag is equal to 0,  ` `    ``// it means there is no answer.  ` `    ``if` `(flag == ``0``)  ` `        ``System.out.println(``"No such subarray!"``); ` `         `  `    ``else` `    ``{  ` `        ``for``(``int` `i = start; i <= end; i++)  ` `            ``System.out.print(arr[i] + ``" "``); ` `    ``}  ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``2``, ``3``, ``2``, ``4``, ``5` `};  ` `    ``int` `n = arr.length;  ` ` `  `    ``FindSubarray(arr, n);  ` `} ` `} ` ` `  `// This code is contributed by divyeshrabadiya07 `

## Python3

 `# Python3 program for the above approach ` `import` `sys ` ` `  `# Function to find subarray ` `def` `FindSubarray(arr, n): ` ` `  `    ``# If the array has only one element, ` `    ``# then there is no answer. ` `    ``if` `(n ``=``=` `1``): ` `        ``print``(``"No such subarray!"``) ` ` `  `    ``# Array to store the last occurrences ` `    ``# of the elements of the array. ` `    ``vis ``=` `[``-``1``] ``*` `(n ``+` `1``) ` `    ``vis[arr[``0``]] ``=` `0` ` `  `    ``# To maintain the length ` `    ``length ``=` `sys.maxsize ` `    ``flag ``=` `0` `     `  `    ``for` `i ``in` `range``(``1``, n): ` `        ``t ``=` `arr[i] ` ` `  `        ``# Check if element is occurring ` `        ``# for the second time in the array ` `        ``if` `(vis[t] !``=` `-``1``): ` `             `  `            ``# Find distance between last ` `            ``# and current index ` `            ``# of the element. ` `            ``distance ``=` `i ``-` `vis[t] ``+` `1` ` `  `            ``# If the current distance ` `            ``# is less then len ` `            ``# update len and ` `            ``# set 'start' and 'end' ` `            ``if` `(distance < length): ` `                ``length ``=` `distance ` `                ``start ``=` `vis[t] ` `                ``end ``=` `i ` `             `  `            ``flag ``=` `1` ` `  `        ``# Set the last occurrence ` `        ``# of current element to be 'i'. ` `        ``vis[t] ``=` `i ` ` `  `    ``# If flag is equal to 0, ` `    ``# it means there is no answer. ` `    ``if` `(flag ``=``=` `0``): ` `        ``print``(``"No such subarray!"``) ` `    ``else``: ` `        ``for` `i ``in` `range``(start, end ``+` `1``): ` `            ``print``(arr[i], end ``=` `" "``) ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` ` `  `    ``arr ``=` `[ ``2``, ``3``, ``2``, ``4``, ``5` `] ` `    ``n ``=` `len``(arr) ` ` `  `    ``FindSubarray(arr, n) ` ` `  `# This code is contributed by chitranayal `

## C#

 `// C# program for the above approach ` `using` `System; ` `class` `GFG{ ` `     `  `// Function to find subarray  ` `public` `static` `void` `FindSubarray(``int``[] arr,  ` `                                ``int` `n)  ` `{  ` `     `  `    ``// If the array has only one element,  ` `    ``// then there is no answer.  ` `    ``if` `(n == 1) ` `    ``{  ` `        ``Console.WriteLine(``"No such subarray!"``); ` `    ``}  ` ` `  `    ``// Array to store the last occurrences  ` `    ``// of the elements of the array.  ` `    ``int``[] vis = ``new` `int``[n + 1];  ` `    ``for``(``int` `i = 0; i < n + 1; i++) ` `        ``vis[i] = -1; ` `    ``vis[arr] = 0;  ` ` `  `    ``// To maintain the length  ` `    ``int` `len = ``int``.MaxValue, flag = 0;  ` ` `  `    ``// Variables to store  ` `    ``// start and end indices  ` `    ``int` `start = 0, end = 0;  ` ` `  `    ``for``(``int` `i = 1; i < n; i++) ` `    ``{  ` `        ``int` `t = arr[i];  ` ` `  `        ``// Check if element is occurring  ` `        ``// for the second time in the array  ` `        ``if` `(vis[t] != -1) ` `        ``{ ` `             `  `            ``// Find distance between last  ` `            ``// and current index  ` `            ``// of the element.  ` `            ``int` `distance = i - vis[t] + 1;  ` ` `  `            ``// If the current distance  ` `            ``// is less then len  ` `            ``// update len and  ` `            ``// set 'start' and 'end'  ` `            ``if` `(distance < len) ` `            ``{  ` `                ``len = distance;  ` `                ``start = vis[t];  ` `                ``end = i;  ` `            ``}  ` `            ``flag = 1;  ` `        ``}  ` ` `  `        ``// Set the last occurrence  ` `        ``// of current element to be 'i'.  ` `        ``vis[t] = i;  ` `    ``}  ` `     `  `    ``// If flag is equal to 0,  ` `    ``// it means there is no answer.  ` `    ``if` `(flag == 0)  ` `        ``Console.WriteLine(``"No such subarray!"``); ` `         `  `    ``else` `    ``{  ` `        ``for``(``int` `i = start; i <= end; i++)  ` `            ``Console.Write(arr[i] + ``" "``); ` `    ``}  ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]arr = { 2, 3, 2, 4, 5 };  ` `    ``int` `n = arr.Length;  ` ` `  `    ``FindSubarray(arr, n);  ` `} ` `} ` ` `  `// This code is contributed by sapnasingh4991`

Output:

```2 3 2
```

Time Complexity: O(N), where n is the length of the array.

Auxiliary Space: O(N).

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