Given an array arr[] having N integers in non-decreasing order and an integer K, the task is to find the minimum value of (j – i) for a pair (i, j) such that the range [arr[i], arr[j]] contains at least K odd integers.
Examples:
Input: arr[] = {1, 3, 6, 8, 15, 21}, K = 3
Output: 1
Explanation: For (i, j) = (3, 4), it represents the range [8, 15] having 4 odd integers {9, 11, 13, 15} (i.e, more than K). Hence the value of j – i = 1, which is the minimum possible.Input: arr[] = {5, 6, 7, 8, 9}, K = 5
Output: -1
Approach: The given problem can be solved using the two-pointers approach and some basic mathematics. The idea is to use a sliding window to check the count of odd numbers in the range and find the size of the smallest window containing at least K odd numbers. It can be done by maintaining two pointers i and j and calculating the count of odd numbers in the range [arr[i], arr[j]]. Maintain the minimum value of (j – i) in a variable for windows with more than K odd integers which is the required answer.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the minimum value of j - i// such that the count of odd integers in the// range [arr[i], arr[j]] is more than Kint findEven(int arr[], int N, int K){ // Base Case int L = arr[0]; int R = arr[N - 1]; // Maximum count of odd integers int Count = (L & 1) ? ceil((float)(R - L + 1) / 2) : (R - L + 1) / 2; // If no valid (i, j) exists if (K > Count) { return -1; } // Initialize the variables int i = 0, j = 0, ans = INT_MAX; // Loop for the two pointer approach while (j < N) { L = arr[i]; R = arr[j]; // Calculate count of odd numbers in the // range [L, R] Count = (L & 1) ? ceil((float)(R - L + 1) / 2) : (R - L + 1) / 2; if (K > Count) { j++; } else { // If the current value of j - i // is smaller, update answer if (j - i < ans) { ans = j - i; } i++; } } // Return Answer return ans;}// Driver Codeint main(){ int arr[] = { 1, 3, 6, 8, 15, 21 }; int N = sizeof(arr) / sizeof(arr[0]); int K = 3; cout << findEven(arr, N, K); return 0;} |
Java
// Java program for the above approachimport java.util.*;public class GFG{ // Function to find the minimum value of j - i// such that the count of odd integers in the// range [arr[i], arr[j]] is more than Kstatic int findEven(int []arr, int N, int K){ // Base Case int L = arr[0]; int R = arr[N - 1]; int Count = 0; // Maximum count of odd integers if ((L & 1) > 0) { Count = (int)Math.ceil((float)(R - L + 1) / 2.0); } else { Count = (R - L + 1) / 2; } // If no valid (i, j) exists if (K > Count) { return -1; } // Initialize the variables int i = 0, j = 0, ans = Integer.MAX_VALUE; // Loop for the two pointer approach while (j < N) { L = arr[i]; R = arr[j]; // Calculate count of odd numbers in the // range [L, R] if ((L & 1) > 0) { Count = (int)Math.ceil((float)(R - L + 1) / 2); } else { Count = (R - L + 1) / 2; } if (K > Count) { j++; } else { // If the current value of j - i // is smaller, update answer if (j - i < ans) { ans = j - i; } i++; } } // Return Answer return ans;}// Driver Codepublic static void main(String args[]){ int []arr = { 1, 3, 6, 8, 15, 21 }; int N = arr.length; int K = 3; System.out.println(findEven(arr, N, K));}}// This code is contributed by Samim Hossain Mondal. |
Python3
# Python Program to implement# the above approachimport math as Math# Function to find the minimum value of j - i# such that the count of odd integers in the# range [arr[i], arr[j]] is more than Kdef findEven(arr, N, K): # Base Case L = arr[0] R = arr[N - 1] # Maximum count of odd integers Count = Math.ceil((R - L + 1) / 2) if (L & 1) else (R - L + 1) / 2 # If no valid (i, j) exists if (K > Count): return -1 # Initialize the variables i = 0 j = 0 ans = 10**9 # Loop for the two pointer approach while (j < N): L = arr[i] R = arr[j] # Calculate count of odd numbers in the # range [L, R] Count = Math.ceil((R - L + 1) / 2) if (L & 1) else (R - L + 1) / 2 if (K > Count): j += 1 else: # If the current value of j - i # is smaller, update answer if (j - i < ans): ans = j - i i += 1 # Return Answer return ans# Driver Codearr = [1, 3, 6, 8, 15, 21]N = len(arr)K = 3print(findEven(arr, N, K))# This code is contributed by gfgking |
C#
// C# program for the above approachusing System;public class GFG{ // Function to find the minimum value of j - i// such that the count of odd integers in the// range [arr[i], arr[j]] is more than Kstatic int findEven(int []arr, int N, int K){ // Base Case int L = arr[0]; int R = arr[N - 1]; int Count = 0; // Maximum count of odd integers if ((L & 1) > 0) { Count = (int)Math.Ceiling((float)(R - L + 1) / 2.0); } else { Count = (R - L + 1) / 2; } // If no valid (i, j) exists if (K > Count) { return -1; } // Initialize the variables int i = 0, j = 0, ans = Int32.MaxValue; // Loop for the two pointer approach while (j < N) { L = arr[i]; R = arr[j]; // Calculate count of odd numbers in the // range [L, R] if ((L & 1) > 0) { Count = (int)Math.Ceiling((float)(R - L + 1) / 2); } else { Count = (R - L + 1) / 2; } if (K > Count) { j++; } else { // If the current value of j - i // is smaller, update answer if (j - i < ans) { ans = j - i; } i++; } } // Return Answer return ans;}// Driver Codepublic static void Main(){ int []arr = { 1, 3, 6, 8, 15, 21 }; int N = arr.Length; int K = 3; Console.Write(findEven(arr, N, K));}}// This code is contributed by Samim Hossain Mondal. |
Javascript
<script> // JavaScript Program to implement // the above approach // Function to find the minimum value of j - i // such that the count of odd integers in the // range [arr[i], arr[j]] is more than K function findEven(arr, N, K) { // Base Case let L = arr[0]; let R = arr[N - 1]; // Maximum count of odd integers let Count = (L & 1) ? Math.ceil((R - L + 1) / 2) : (R - L + 1) / 2; // If no valid (i, j) exists if (K > Count) { return -1; } // Initialize the variables let i = 0, j = 0, ans = Number.MAX_VALUE; // Loop for the two pointer approach while (j < N) { L = arr[i]; R = arr[j]; // Calculate count of odd numbers in the // range [L, R] Count = (L & 1) ? Math.ceil((R - L + 1) / 2) : (R - L + 1) / 2; if (K > Count) { j++; } else { // If the current value of j - i // is smaller, update answer if (j - i < ans) { ans = j - i; } i++; } } // Return Answer return ans; } // Driver Code let arr = [1, 3, 6, 8, 15, 21]; let N = arr.length; let K = 3; document.write(findEven(arr, N, K)); // This code is contributed by Potta Lokesh </script> |
Output
1
Time Complexity: O(N)
Auxiliary Space: O(1)
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