Find the longest sub-array having exactly k odd numbers
Last Updated :
02 Aug, 2022
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 numbers
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 to test above
int 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 implementation to find the longest
// sub-array having exactly k odd numbers
import 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 implementation to find the longest
# sub-array having exactly k odd numbers
# Function to find the longest sub-array
# having exactly k odd numbers
def 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 Code
arr = [3, 4, 6, 1, 9, 8, 2, 10]
n = len(arr)
k = 2
print("Length = ", longSubarrWithKOddNum(arr, n, k))
# This code is contributed by Nikita Tiwari.
// C# implementation to find the longest
// sub-array having exactly k odd numbers
using 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 implementation to find the longest
// sub-array having exactly k odd numbers
// function to find the longest sub-array
// having exactly k odd numbers
function 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.
?>
<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 numbers
function 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 above
var arr = [3, 4, 6, 1, 9, 8, 2, 10];
var n = arr.length;
var k = 2;
document.write( "Length = "
+ longSubarrWithKOddNum(arr, n, k));
</script>
Time Complexity: O(n).
Similar Reads
DSA Tutorial - Learn Data Structures and Algorithms DSA (Data Structures and Algorithms) is the study of organizing data efficiently using data structures like arrays, stacks, and trees, paired with step-by-step procedures (or algorithms) to solve problems effectively. Data structures manage how data is stored and accessed, while algorithms focus on
7 min read
Quick Sort QuickSort is a sorting algorithm based on the Divide and Conquer that picks an element as a pivot and partitions the given array around the picked pivot by placing the pivot in its correct position in the sorted array. It works on the principle of divide and conquer, breaking down the problem into s
12 min read
Merge Sort - Data Structure and Algorithms Tutorials Merge sort is a popular sorting algorithm known for its efficiency and stability. It follows the divide-and-conquer approach. It works by recursively dividing the input array into two halves, recursively sorting the two halves and finally merging them back together to obtain the sorted array. Merge
14 min read
Data Structures Tutorial Data structures are the fundamental building blocks of computer programming. They define how data is organized, stored, and manipulated within a program. Understanding data structures is very important for developing efficient and effective algorithms. What is Data Structure?A data structure is a st
2 min read
Bubble Sort Algorithm Bubble Sort is the simplest sorting algorithm that works by repeatedly swapping the adjacent elements if they are in the wrong order. This algorithm is not suitable for large data sets as its average and worst-case time complexity are quite high.We sort the array using multiple passes. After the fir
8 min read
Breadth First Search or BFS for a Graph Given a undirected graph represented by an adjacency list adj, where each adj[i] represents the list of vertices connected to vertex i. Perform a Breadth First Search (BFS) traversal starting from vertex 0, visiting vertices from left to right according to the adjacency list, and return a list conta
15+ min read
Binary Search Algorithm - Iterative and Recursive Implementation Binary Search Algorithm is a searching algorithm used in a sorted array by repeatedly dividing the search interval in half. The idea of binary search is to use the information that the array is sorted and reduce the time complexity to O(log N). Binary Search AlgorithmConditions to apply Binary Searc
15 min read
Insertion Sort Algorithm Insertion sort is a simple sorting algorithm that works by iteratively inserting each element of an unsorted list into its correct position in a sorted portion of the list. It is like sorting playing cards in your hands. You split the cards into two groups: the sorted cards and the unsorted cards. T
9 min read
Array Data Structure Guide In this article, we introduce array, implementation in different popular languages, its basic operations and commonly seen problems / interview questions. An array stores items (in case of C/C++ and Java Primitive Arrays) or their references (in case of Python, JS, Java Non-Primitive) at contiguous
4 min read
Sorting Algorithms A Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
3 min read