# Minimum length of subarray in given Ternary Array having 0 as the majority element

• Last Updated : 23 Dec, 2021

Given an integer array arr[] of size n with only three types of integers 0’s, 1’s, and 2’s. Find the minimum length of the subarray of the array arr[] of length >=2, such that it has the frequency of 0’s greater than both 1’s and 2’s. If not found print -1

Input : arr[] = {2, 0, 2, 0, 1, 2, 2, 2}
Output : 3
Explanation: {0, 2, 0} from index [2, 4] is the subarray with frequencies of 0’s greater frequencies of 1’s and 2’s.

Input: arr[] = {2, 2, 2, 2}
Output: -1

Naive Approach: This problem can be done by going through every subarray and checking the frequencies of 0’s, 1’s and 2’s and checking if the frequency of 0’s is greater than both 2’s and 1’s then storing the minimum length of subarray as the answer.

Time Complexity: O(n*n)
Auxiliary Space: O(1)

Efficient Approach: Since there are only 3 types of integers the possible subarrays of length >=2 satisfying the above condition would be

{0, 0}, {0, 1, 0}, {0, 2, 0}, {0, 1, 2, 0}, {0, 2, 1, 0}, {0, 0, 0, 1, 1, 2, 2}, ….

The maximum possible minimum length of subarray would be 7. Any other subarray which satisfies the above condition with length > 7 would contain any of the above subarrays in it so the maximum possible length of subarray which satisfies the above condition is 7.

Follow these steps to solve this problem:

• Initialize a variable min_length as INT_MAX
• Iterate in the range [0, n) using the variable i and perform the following tasks:
• Initialize an array count[] with initial frequencies 0
• Increment the frequency of arr[i] using count[arr[i]]++.
• Iterate in the range [i+1, min(n, i+7)) using the variable j and perform the following tasks:
• Increment the frequency of arr[j] by using count[arr[j]]++ of each subarray of size <=7.
• If the count is greater than both count and count and if min_length > j-i+1 then assign min_length = j-i+1
• If min_length is equal to INT_MAX return -1.
• Else print min_length as the answer.

Below is the implementation of the above approach.

## C++

 `// C++ code for the above approach``#include ``using` `namespace` `std;` `// Function to find the minimum``// subarray length with most 0's``int` `minLength(vector<``int``> arr)``{``    ``int` `min_len = INT_MAX;``    ``int` `n = arr.size();` `    ``// Traverse the array to check``    ``// the required condition``    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// Initialize a vector count``        ``// to store the frequencies``        ``int` `count = { 0 };``        ``count[arr[i]]++;` `        ``// Check all subarrays of length``        ``// <=7 and count the frequencies``        ``for` `(``int` `j = i + 1; j < min(n, i + 7); j++) {``            ``count[arr[j]]++;` `            ``// If the frequency of 0's is``            ``// greater than both 1's and 2's``            ``// then take length of minimum subarray``            ``if` `(count > count``                ``&& count > count)``                ``min_len = min(min_len, j - i + 1);``        ``}``    ``}` `    ``// If min_len == INT_MAX we have no subarray``    ``// satisfying the condition return -1;``    ``if` `(min_len == INT_MAX)``        ``min_len = -1;` `    ``return` `min_len;``}` `// Driver Code``int` `main()``{` `    ``// Initializing list of nums``    ``vector<``int``> arr = { 2, 0, 2, 0, 1, 2, 2, 2 };` `    ``// Call the function and print the answer``    ``cout << (minLength(arr));` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;` `class` `GFG {` `    ``// Function to find the minimum``    ``// subarray length with most 0's``    ``static` `int` `minLength(ArrayList arr)``    ``{``        ``int` `min_len = Integer.MAX_VALUE;``        ``int` `n = arr.size();` `        ``// Traverse the array to check``        ``// the required condition``        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``// Initialize a vector count``            ``// to store the frequencies``            ``int``[] count = ``new` `int``[``3``];``            ``for` `(``int` `j = ``0``; j < ``3``; j++) {``                ``count[j] = ``0``;``            ``}` `            ``int` `x = (``int``)arr.get(i);``            ``count[x]++;` `            ``// Check all subarrays of length``            ``// <=7 and count the frequencies``            ``for` `(``int` `j = i + ``1``; j < Math.min(n, i + ``7``);``                 ``j++) {``                ``int` `y = (``int``)arr.get(j);``                ``count[y]++;` `                ``// If the frequency of 0's is``                ``// greater than both 1's and 2's``                ``// then take length of minimum subarray``                ``if` `(count[``0``] > count[``1``]``                    ``&& count[``0``] > count[``2``])``                    ``min_len = Math.min(min_len, j - i + ``1``);``            ``}``        ``}` `        ``// If min_len == INT_MAX we have no subarray``        ``// satisfying the condition return -1;``        ``if` `(min_len == Integer.MAX_VALUE)``            ``min_len = -``1``;` `        ``return` `min_len;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``ArrayList arr = ``new` `ArrayList();` `        ``arr.add(``2``);``        ``arr.add(``0``);``        ``arr.add(``2``);``        ``arr.add(``0``);``        ``arr.add(``1``);``        ``arr.add(``2``);``        ``arr.add(``2``);``        ``arr.add(``2``);` `        ``// Count of isograms in string array arr[]``        ``System.out.println(minLength(arr));``    ``}``}` `// This code is contributed by ukasp.`

## Python3

 `# python code for the above approach``INT_MAX ``=` `2147483647` `# Function to find the minimum``# subarray length with most 0's``def` `minLength(arr):` `    ``min_len ``=` `INT_MAX``    ``n ``=` `len``(arr)` `    ``# Traverse the array to check``    ``# the required condition``    ``for` `i ``in` `range``(``0``, n):` `        ``# Initialize a vector count``        ``# to store the frequencies``        ``count ``=` `[``0``, ``0``, ``0``]``        ``count[arr[i]] ``+``=` `1` `        ``# Check all subarrays of length``        ``# <=7 and count the frequencies``        ``for` `j ``in` `range``(i``+``1``, ``min``(n, i``+``7``)):``            ``count[arr[j]] ``+``=` `1` `            ``# If the frequency of 0's is``            ``# greater than both 1's and 2's``            ``# then take length of minimum subarray``            ``if` `(count[``0``] > count[``1``] ``and` `count[``0``] > count[``2``]):``                ``min_len ``=` `min``(min_len, j ``-` `i ``+` `1``)` `    ``# If min_len == INT_MAX we have no subarray``    ``# satisfying the condition return -1;``    ``if` `(min_len ``=``=` `INT_MAX):``        ``min_len ``=` `-``1` `    ``return` `min_len` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `    ``# Initializing list of nums``    ``arr ``=` `[``2``, ``0``, ``2``, ``0``, ``1``, ``2``, ``2``, ``2``]` `    ``# Call the function and print the answer``    ``print``(minLength(arr))` `    ``# This code is contributed by rakeshsahni`

## C#

 `// C# program for the above approach``using` `System;``using` `System.Collections;` `class` `GFG{` `// Function to find the minimum``// subarray length with most 0's``static` `int` `minLength(ArrayList arr)``{``    ``int` `min_len = Int32.MaxValue;``    ``int` `n = arr.Count;` `    ``// Traverse the array to check``    ``// the required condition``    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// Initialize a vector count``        ``// to store the frequencies``        ``int` `[]count = ``new` `int``;``        ``for``(``int` `j = 0; j < 3; j++) {``            ``count[j] = 0;``        ``}``        ` `        ``int` `x = (``int``)arr[i];``        ``count[x]++;` `        ``// Check all subarrays of length``        ``// <=7 and count the frequencies``        ``for` `(``int` `j = i + 1; j < Math.Min(n, i + 7); j++) {``          ``int` `y = (``int``)arr[j];``            ``count[y]++;` `            ``// If the frequency of 0's is``            ``// greater than both 1's and 2's``            ``// then take length of minimum subarray``            ``if` `(count > count``                ``&& count > count)``                ``min_len =   Math.Min(min_len, j - i + 1);``        ``}``    ``}` `    ``// If min_len == INT_MAX we have no subarray``    ``// satisfying the condition return -1;``    ``if` `(min_len == Int32.MaxValue)``        ``min_len = -1;` `    ``return` `min_len;``}` `// Driver Code``public` `static` `void` `Main()``{``    ``ArrayList arr = ``new` `ArrayList();``    ` `    ``arr.Add(2);``    ``arr.Add(0);``    ``arr.Add(2);``    ``arr.Add(0);``    ``arr.Add(1);``    ``arr.Add(2);``    ``arr.Add(2);``    ``arr.Add(2);` `    ``// Count of isograms in string array arr[]``    ``Console.WriteLine(minLength(arr));``}``}` `// This code is contributed by Samim Hossain Mondal.`

## Javascript

 ``
Output
`3`

Time Complexity: O(n)
Auxiliary Space: O(1)

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