Given a range [low, high], both inclusive, and an integer K, the task is to select K numbers from the range(a number can be chosen multiple times) such that the sum of those K numbers is even. Print the number of all such permutations.
Examples:
Input: low = 4, high = 5, k = 3
Output: 4
Explanation:
There are 4 valid permutation. They are {4, 4, 4}, {4, 5, 5}, {5, 4, 5} and {5, 5, 4} which sum up to an even number.Input: low = 1, high = 10, k = 2
Output: 50
Explanation:
There are 50 valid permutations. They are {1, 1}, {1, 3}, .. {1, 9} {2, 2}, {2, 4}, …, {2, 10}, …, {10, 2}, {10, 4}, … {10, 10}.
These 50 permutations, each sum up to an even number.
Naive Approach: The idea is to find all subset of size K such that the sum of the subset is even and also calculate permutation for each required subset.
Time Complexity: O(K * (2K))
Auxiliary Space: O(K)
Efficient Approach: The idea is to use the fact that the sum of two even and odd numbers is always even. Follow the steps below to solve the problem:
- Find the total count of even and odd numbers in the given range [low, high].
- Initialize variable even_sum = 1 and odd_sum = 0 to store way to get even sum and odd sum respectively.
- Iterate a loop K times and store the previous even sum as prev_even = even_sum and the previous odd sum as prev_odd = odd_sum where even_sum = (prev_even*even_count) + (prev_odd*odd_count) and odd_sum = (prev_even*odd_count) + (prev_odd*even_count).
- Print the even_sum at the end as there is a count for the odd sum because the previous odd_sum will contribute to the next even_sum.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include<bits/stdc++.h>using namespace std;// Function to return the number // of all permutations such that // sum of K numbers in range is even int countEvenSum(int low, int high, int k) { // Find total count of even and // odd number in given range int even_count = high / 2 - (low - 1) / 2; int odd_count = (high + 1) / 2 - low / 2; long even_sum = 1; long odd_sum = 0; // Iterate loop k times and update // even_sum & odd_sum using // previous values for(int i = 0; i < k; i++) { // Update the prev_even and // odd_sum long prev_even = even_sum; long prev_odd = odd_sum; // Even sum even_sum = (prev_even * even_count) + (prev_odd * odd_count); // Odd sum odd_sum = (prev_even * odd_count) + (prev_odd * even_count); } // Return even_sum cout << (even_sum); } // Driver Code int main(){ // Given ranges int low = 4; int high = 5; // Length of permutation int K = 3; // Function call countEvenSum(low, high, K); }// This code is contributed by Stream_Cipher |
Java
// Java program for the above approachimport java.util.*;class GFG { // Function to return the number // of all permutations such that // sum of K numbers in range is even public static void countEvenSum(int low, int high, int k) { // Find total count of even and // odd number in given range int even_count = high / 2 - (low - 1) / 2; int odd_count = (high + 1) / 2 - low / 2; long even_sum = 1; long odd_sum = 0; // Iterate loop k times and update // even_sum & odd_sum using // previous values for (int i = 0; i < k; i++) { // Update the prev_even and // odd_sum long prev_even = even_sum; long prev_odd = odd_sum; // Even sum even_sum = (prev_even * even_count) + (prev_odd * odd_count); // Odd sum odd_sum = (prev_even * odd_count) + (prev_odd * even_count); } // Return even_sum System.out.println(even_sum); } // Driver Code public static void main(String[] args) { // Given ranges int low = 4; int high = 5; // Length of permutation int K = 3; // Function call countEvenSum(low, high, K); }} |
Python3
# Python3 program for the above approach # Function to return the number # of all permutations such that # sum of K numbers in range is even def countEvenSum(low, high, k): # Find total count of even and # odd number in given range even_count = high / 2 - (low - 1) / 2 odd_count = (high + 1) / 2 - low / 2 even_sum = 1 odd_sum = 0 # Iterate loop k times and update # even_sum & odd_sum using # previous values for i in range(0, k): # Update the prev_even and # odd_sum prev_even = even_sum prev_odd = odd_sum # Even sum even_sum = ((prev_even * even_count) + (prev_odd * odd_count)) # Odd sum odd_sum = ((prev_even * odd_count) + (prev_odd * even_count)) # Return even_sum print(int(even_sum)) # Driver Code # Given ranges low = 4; high = 5; # Length of permutation K = 3; # Function call countEvenSum(low, high, K); # This code is contributed by Stream_Cipher |
C#
// C# program for the above approachusing System;class GFG{// Function to return the number// of all permutations such that// sum of K numbers in range is evenpublic static void countEvenSum(int low, int high, int k){ // Find total count of even and // odd number in given range int even_count = high / 2 - (low - 1) / 2; int odd_count = (high + 1) / 2 - low / 2; long even_sum = 1; long odd_sum = 0; // Iterate loop k times and update // even_sum & odd_sum using // previous values for(int i = 0; i < k; i++) { // Update the prev_even and // odd_sum long prev_even = even_sum; long prev_odd = odd_sum; // Even sum even_sum = (prev_even * even_count) + (prev_odd * odd_count); // Odd sum odd_sum = (prev_even * odd_count) + (prev_odd * even_count); } // Return even_sum Console.WriteLine(even_sum);}// Driver Codepublic static void Main(String[] args){ // Given ranges int low = 4; int high = 5; // Length of permutation int K = 3; // Function call countEvenSum(low, high, K);}}// This code is contributed by amal kumar choubey |
Javascript
<script>// JavaScript program for the above approach // Function to return the number // of all permutations such that // sum of K numbers in range is even function countEvenSum(low, high, k) { // Find total count of even and // odd number in given range let even_count = high / 2 - (low - 1) / 2; let odd_count = (high + 1) / 2 - low / 2; let even_sum = 1; let odd_sum = 0; // Iterate loop k times and update // even_sum & odd_sum using // previous values for (let i = 0; i < k; i++) { // Update the prev_even and // odd_sum let prev_even = even_sum; let prev_odd = odd_sum; // Even sum even_sum = (prev_even * even_count) + (prev_odd * odd_count); // Odd sum odd_sum = (prev_even * odd_count) + (prev_odd * even_count); } // Return even_sum document.write(even_sum); }// Driver Code // Given ranges let low = 4; let high = 5; // Length of permutation let K = 3; // Function call countEvenSum(low, high, K); </script> |
Output:
4
Time Complexity: O(K)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
