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 a 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.
- Iterate over the given array arr[]
- Find the number of zeros, one’s and two’s in the current subarray
- Check if the current subarray size is greater than equals to 2 and the number of zeros is greater than the number of ones and two’s then, update the min_len
- Check if a valid subarray exists.
- if not, return -1
- Return min_len;
Below is the implementation of the above approach:
C++
// C++ code for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the minimum// subarray length with most 0'sint minLength(vector<int> arr){ int min_len = INT_MAX; int n = arr.size(); // Iterate over the given array arr[] for (int i = 0; i < n; i++) { int zeroCount = 0; int oneCount = 0; int twoCount = 0; // Find the number of zero, one and two in the // current subarray for (int j = i; j < n; j++) { if (arr[j] == 0) zeroCount++; else if (arr[j] == 1) oneCount++; else if (arr[j] == 2) twoCount++; int currSubarrayLength = (j - i + 1); // Check if the current subarray size is greater // than equals to 2 and number of zero is // greater that number of one's and two's then, // update the min_len if (currSubarrayLength >= 2 && zeroCount > oneCount && zeroCount > twoCount) { min_len = min(min_len, (j - i + 1)); } } } // Check if valid subarray exist. // if not, return -1 if (min_len == INT_MAX) return -1; // Return min_len; return min_len;}// Driver Codeint 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
import java.util.Arrays;import java.util.List;public class Gfg { // Function to find the minimum // subarray length with most 0's static int minLength(List<Integer> arr) { int minLen = Integer.MAX_VALUE; int n = arr.size(); // Iterate over the given array arr[] for (int i = 0; i < n; i++) { int zeroCount = 0; int oneCount = 0; int twoCount = 0; // Find the number of zero, one and two in the // current subarray for (int j = i; j < n; j++) { if (arr.get(j) == 0) zeroCount++; else if (arr.get(j) == 1) oneCount++; else if (arr.get(j) == 2) twoCount++; int currSubarrayLength = (j - i + 1); // Check if the current subarray size is // greater than equals to 2 and number of // zero is greater that number of one's and // two's then, update the min_len if (currSubarrayLength >= 2 && zeroCount > oneCount && zeroCount > twoCount) { minLen = Math.min(minLen, (j - i + 1)); } } } // Check if valid subarray exist. // if not, return -1 if (minLen == Integer.MAX_VALUE) return -1; // Return minLen; return minLen; } public static void main(String[] args) { // Initializing list of nums List<Integer> arr = Arrays.asList(2, 0, 2, 0, 1, 2, 2, 2); // Call the function and print the answer System.out.println(minLength(arr)); }} |
Python3
# Python code implementation for the above approach# Function to find the minimum# subarray length with most 0'sdef minLength(arr): minLen = float('inf') n = len(arr) # Iterate over the given array arr[] for i in range(n): zeroCount = 0 oneCount = 0 twoCount = 0 # Find the number of zero, one # and two in the current subarray for j in range(i, n): if arr[j] == 0: zeroCount += 1 elif arr[j] == 1: oneCount += 1 elif arr[j] == 2: twoCount += 1 currSubarrayLength = (j - i + 1) # Check if the current subarray size is # greater than equals to 2 and number of # zero is greater that number of one's and # two's then, update the min_len if currSubarrayLength >= 2 and zeroCount > oneCount and zeroCount > twoCount: minLen = min(minLen, (j - i + 1)) # Check if valid subarray exist. # if not, return -1 if minLen == float('inf'): return -1 # Return minLen return minLen # Initializing list of numsarr = [2, 0, 2, 0, 1, 2, 2, 2]# Call the function and print the answerprint(minLength(arr))# This code is contributed by lokeshmvs21. |
Javascript
// Function to find the minimum subarray length with most 0'sfunction minLength(arr) { let min_len = Number.MAX_SAFE_INTEGER; let n = arr.length; // Iterate over the given array arr[] for (let i = 0; i < n; i++) { let zeroCount = 0; let oneCount = 0; let twoCount = 0; // Find the number of zero, one and two in the current subarray for (let j = i; j < n; j++) { if (arr[j] === 0) { zeroCount++; } else if (arr[j] === 1) { oneCount++; } else if (arr[j] === 2) { twoCount++; } let currSubarrayLength = j - i + 1; // Check if the current subarray size is greater than equals to 2 // and number of zero is greater than number of one's and two's // then, update the min_len if ( currSubarrayLength >= 2 && zeroCount > oneCount && zeroCount > twoCount ) { min_len = Math.min(min_len, j - i + 1); } }}// Check if valid subarray exist. If not, return -1if (min_len === Number.MAX_SAFE_INTEGER) {return -1;}// Return min_len;return min_len;}// Driver Codelet arr = [2, 0, 2, 0, 1, 2, 2, 2];// Call the function and print the answerconsole.log(minLength(arr)); |
C#
using System;using System.Linq;using System.Collections.Generic;class Gfg { // Function to find the minimum // subarray length with most 0's static int minLength(List<int> arr) { int minLen = int.MaxValue; int n = arr.Count; // Iterate over the given array arr[] for (int i = 0; i < n; i++) { int zeroCount = 0; int oneCount = 0; int twoCount = 0; // Find the number of zero, one and two in the // current subarray for (int j = i; j < n; j++) { if (arr[j] == 0) zeroCount++; else if (arr[j] == 1) oneCount++; else if (arr[j] == 2) twoCount++; int currSubarrayLength = (j - i + 1); // Check if the current subarray size is // greater than equals to 2 and number of // zero is greater that number of one's and // two's then, update the min_len if (currSubarrayLength >= 2 && zeroCount > oneCount && zeroCount > twoCount) { minLen = Math.Min(minLen, (j - i + 1)); } } } // Check if valid subarray exist. // if not, return -1 if (minLen == int.MaxValue) return -1; // Return minLen; return minLen; } public static void Main(string[] args) { // Initializing list of nums List<int> arr = new List<int>{ 2, 0, 2, 0, 1, 2, 2, 2 }; // Call the function and print the answer Console.WriteLine(minLength(arr)); }} |
Output
3
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[0] is greater than both count[1] and count[2] 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 <bits/stdc++.h>using namespace std;// Function to find the minimum// subarray length with most 0'sint 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[3] = { 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[0] > count[1] && 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 Codeint 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 approachimport 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<Integer> arr = new ArrayList<Integer>(); 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 approachINT_MAX = 2147483647# Function to find the minimum# subarray length with most 0'sdef 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 Codeif __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 approachusing System;using System.Collections;class GFG{// Function to find the minimum// subarray length with most 0'sstatic 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[3]; 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[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 == Int32.MaxValue) min_len = -1; return min_len;}// Driver Codepublic 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
<script> // JavaScript code for the above approach const INT_MAX = 2147483647 // Function to find the minimum // subarray length with most 0's const minLength = (arr) => { let min_len = INT_MAX; let n = arr.length; // Traverse the array to check // the required condition for (let i = 0; i < n; i++) { // Initialize a vector count // to store the frequencies let count = [0, 0, 0]; count[arr[i]]++; // Check all subarrays of length // <=7 and count the frequencies for (let j = i + 1; j < Math.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[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 == INT_MAX) min_len = -1; return min_len; } // Driver Code // Initializing list of nums let arr = [2, 0, 2, 0, 1, 2, 2, 2]; // Call the function and print the answer document.write(minLength(arr)); // This code is contributed by rakeshsahni</script> |
Output
3
Time Complexity: O(n)
Auxiliary Space: O(1)
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