Given an array of size n. The problem is to find the longest sub-array having exactly k odd numbers.
Examples:
Input : arr[] = {2, 3, 4, 11, 4, 12, 7}, k = 1
Output : 4
The sub-array is {4, 11, 4, 12}.
Input : arr[] = {3, 4, 6, 1, 9, 8, 2, 10}, k = 2
Output : 7
The sub-array is {4, 6, 1, 9, 8, 2, 10}.
Naive Approach: Consider all the sub-arrays and count the number of odd numbers in them. Return the length of the one having exactly ‘k’ odd numbers and has the maximum length. Time Complexity is of O(n^2).
Efficient Approach: The idea is to use sliding window. Create a variable count, which stores the number of odd integers in the current window. If the value of count exceeds K at any point, decrease the window size from the start, otherwise include the element in the current window. Similarly, iterate for the complete array and maintain the maximum value of length of all the windows having exactly k odd numbers in a variable max.
longSubarrWithKOddNum(arr, n, k)
Initialize max = 0, count = 0, start = 0
for i = 0 to n-1
if arr[i] % 2 != 0, then
count++
while (count > k && start <= i)
if arr[start++] % 2 != 0, then
count--
if count == k, then
if max < (i - start + 1), then
max = i - start + 1
return max
Implementation:
C++
// C++ implementation to find the longest// sub-array having exactly k odd numbers#include <bits/stdc++.h>using namespace std;// function to find the longest sub-array// having exactly k odd numbersint longSubarrWithKOddNum(int arr[], int n, int k){ int max = 0, count = 0, start = 0; // traverse the given array for (int i = 0; i < n; i++) { // if number is odd increment count if (arr[i] % 2 != 0) count++; // remove elements from sub-array from // 'start' index when count > k while (count > k && start <= i) if (arr[start++] % 2 != 0) count--; // if count == k, then compare max with // current sub-array length if (count == k) if (max < (i - start + 1)) max = i - start + 1; } // required length return max;}// Driver program to test aboveint main(){ int arr[] = {3, 4, 6, 1, 9, 8, 2, 10}; int n = sizeof(arr) / sizeof(arr[0]); int k = 2; cout << "Length = " << longSubarrWithKOddNum(arr, n, k); return 0; } |
Java
// Java implementation to find the longest// sub-array having exactly k odd numbersimport java.io.*;class GFG { // function to find the longest sub-array // having exactly k odd numbers static int longSubarrWithKOddNum(int arr[], int n, int k) { int max = 0, count = 0, start = 0; // traverse the given array for (int i = 0; i < n; i++) { // if number is odd increment count if (arr[i] % 2 != 0) count++; // remove elements from sub-array from // 'start' index when count > k while (count > k && start <= i) if (arr[start++] % 2 != 0) count--; // if count == k, then compare max // with current sub-array length if (count == k) if (max < (i - start + 1)) max = i - start + 1; } // required length return max; } // Driver program public static void main(String args[]) { int arr[] = {3, 4, 6, 1, 9, 8, 2, 10}; int n = arr.length; int k = 2; System.out.println("Length = " + longSubarrWithKOddNum(arr, n, k)); }}// This code is contributed // by Nikita Tiwari. |
Python3
# Python3 implementation to find the longest# sub-array having exactly k odd numbers# Function to find the longest sub-array# having exactly k odd numbersdef longSubarrWithKOddNum(arr, n, k) : mx, count, start = 0, 0, 0 # Traverse the given array for i in range(0, n) : # if number is odd increment count if (arr[i] % 2 != 0) : count = count + 1 # remove elements from sub-array from # 'start' index when count > k while (count > k and start <= i) : if (arr[start] % 2 != 0) : count = count - 1 start = start + 1 # if count == k, then compare max # with current sub-array length if (count == k) : if (mx < (i - start + 1)) : mx = i - start + 1 # required length return mx # Driver Codearr = [3, 4, 6, 1, 9, 8, 2, 10]n = len(arr)k = 2print("Length = ", longSubarrWithKOddNum(arr, n, k))# This code is contributed by Nikita Tiwari. |
C#
// C# implementation to find the longest// sub-array having exactly k odd numbersusing System;class GFG { // function to find the longest sub-array // having exactly k odd numbers static int longSubarrWithKOddNum(int []arr, int n, int k) { int max = 0, count = 0, start = 0; // traverse the given array for (int i = 0; i < n; i++) { // if number is odd increment count if (arr[i] % 2 != 0) count++; // remove elements from sub-array from // 'start' index when count > k while (count > k && start <= i) if (arr[start++] % 2 != 0) count--; // if count == k, then compare max // with current sub-array length if (count == k) if (max < (i - start + 1)) max = i - start + 1; } // required length return max; } // Driver program public static void Main() { int []arr = {3, 4, 6, 1, 9, 8, 2, 10}; int n = arr.Length; int k = 2; Console.WriteLine("Length = " + longSubarrWithKOddNum(arr, n, k)); }}// This code is contributed // by vt_m. |
PHP
<?php// PHP implementation to find the longest// sub-array having exactly k odd numbers// function to find the longest sub-array// having exactly k odd numbersfunction longSubarrWithKOddNum($arr, $n, $k){ $max = 0; $count = 0; $start = 0; // traverse the given array for ($i = 0; $i < $n; $i++) { // if number is odd increment count if ($arr[$i] % 2 != 0) $count++; // remove elements from sub-array from // 'start' index when count > k while ($count > $k && $start <= $i) if ($arr[$start++] % 2 != 0) $count--; // if count == k, then compare max with // current sub-array length if ($count == $k) if ($max < ($i - $start + 1)) $max = $i - $start + 1; } // required length return $max;}// Driver Code{ $arr = array(3, 4, 6, 1, 9, 8, 2, 10); $n = sizeof($arr) / sizeof($arr[0]); $k = 2; echo "Length = ", longSubarrWithKOddNum($arr, $n, $k); return 0; } // This code is contributed by nitin mittal.?> |
Javascript
<script>// Javascript implementation to find the longest// sub-array having exactly k odd numbers// function to find the longest sub-array// having exactly k odd numbersfunction longSubarrWithKOddNum(arr, n, k){ var max = 0, count = 0, start = 0; // traverse the given array for (var i = 0; i < n; i++) { // if number is odd increment count if (arr[i] % 2 != 0) count++; // remove elements from sub-array from // 'start' index when count > k while (count > k && start <= i) if (arr[start++] % 2 != 0) count--; // if count == k, then compare max with // current sub-array length if (count == k) if (max < (i - start + 1)) max = i - start + 1; } // required length return max;}// Driver program to test abovevar arr = [3, 4, 6, 1, 9, 8, 2, 10];var n = arr.length;var k = 2;document.write( "Length = " + longSubarrWithKOddNum(arr, n, k));</script> |
Output
Length = 7
Time Complexity: O(n).
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