Given an array arr[] and brr[] both consisting of N integers and a positive integer K, the task is to find the minimum value of X such that the sum of the maximum of (arr[i] – X, 0) raised to the power of brr[i] for all array elements (arr[i], brr[i]) is at most K.
Examples:
Input: arr[] = {2, 1, 4, 3, 5} brr[] = { 4, 3, 2, 3, 1}, K = 12
Output: 2
Explanation:
Consider the value of X as 2, then the value of the given expression is:
=> max(2 – 2, 0)4 + max(1 – 2, 0)3 + max(4 – 2, 0)2 + max(3 – 2, 0)3 +max(5 – 2, 0)1
=> 04 + 03 + 22 + 13 + 31 = 8 <= K(= 12).
Therefore, the resultant value of X is 2, which is minimum.Input: arr[] = {2, 1, 4, 3, 5} brr[] = { 4, 3, 2, 3, 1}, K = 22
Output: 1
Naive Approach: The simplest approach to solve the given problem is to check for every value of X from 0 to the maximum element of the array and if there exists any value of X satisfying the given conditions, then print that value of X and break out of the loop.
Time Complexity: O(N*M), where, M is the maximum element of the array.
Auxiliary Space: O(1)
Efficient Approach: The above approach can also be optimized by using Binary Search to find the value of X and if a particular value of X satisfies the above condition, then, all the greater values will also satisfy, therefore, then try to search for lower values. Follow the steps below to solve the problem:
- Define a function check(a[], b[], k, n, x):
- Initialize the variable sum as 0 to calculate the desired sum from the array arr[] and brr[].
- Iterate over the range [0, N] using variable i and add the value of pow(max(arr[i] – x, 0), brr[i]) to the variable sum.
- If the value of sum is less than equal to K, then, return true. Otherwise, return false.
- Initialize the variables, say low as 0 and high as maximum value of the array.
- Iterate in a while loop till low is less than high and perform the following steps:
- Initialize the variable mid as the average of low and high.
- Check the value of mid to see whether it satisfies the given conditions by calling the function check(arr[], brr[], k, n, mid).
- If the function check(arr[], brr[], n, k, mid) returns true then, update the high to mid. Otherwise, update the value of low to (mid + 1).
- After completing the above steps, return the value of low as the result from the function.
- After performing the above steps, print the value of low as the desired value of X as the answer.
Below is the implementation of the above approach:
C++14
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to check if there exists an// X that satisfies the given conditionsbool check(int a[], int b[], int k, int n, int x){ int sum = 0; // Find the required value of the // given expression for (int i = 0; i < n; i++) { sum = sum + pow(max(a[i] - x, 0), b[i]); } if (sum <= k) return true; else return false;}// Function to find the minimum value// of X using binary search.int findMin(int a[], int b[], int n, int k){ // Boundaries of the Binary Search int l = 0, u = *max_element(a, a + n); while (l < u) { // Find the middle value int m = (l + u) / 2; // Check for the middle value if (check(a, b, k, n, m)) { // Update the upper u = m; } else { // Update the lower l = m + 1; } } return l;}// Driver Codeint main(){ int arr[] = { 2, 1, 4, 3, 5 }; int brr[] = { 4, 3, 2, 3, 1 }; int K = 12; int N = sizeof(arr) / sizeof(arr[0]); cout << findMin(arr, brr, N, K); return 0;} |
Java
// Java program for the above approach import java.io.*;class GFG{ // Function to check if it is possible to// get desired resultstatic boolean check(int a[], int b[], int k, int x){ int sum = 0; for(int i = 0; i < a.length; i++) { sum = sum + (int)Math.pow( Math.max(a[i] - x, 0), b[i]); } if (sum <= k) return true; else return false;}// Function to find the minimum value // of X using binary search. static int findMin(int a[], int b[], int n, int k){ // Boundaries of the Binary Search int l = 0, u = (int)1e9; while (l < u) { // Find the middle value int m = (l + u) / 2; // Check for the middle value if (check(a, b, k, m)) // Update the upper u = m; else // Update the lower l = m + 1; } return l;}// Driver codepublic static void main(String[] args){ int n = 5; int k = 12; int a[] = { 2, 1, 4, 3, 5 }; int b[] = { 4, 3, 2, 3, 1 }; System.out.println(findMin(a, b, n, k));}}// This code is contributed by ayush_dragneel |
Python3
# Python 3 program for the above approach# Function to check if there exists an# X that satisfies the given conditionsdef check(a, b, k, n, x): sum = 0 # Find the required value of the # given expression for i in range(n): sum = sum + pow(max(a[i] - x, 0), b[i]) if (sum <= k): return True else: return False# Function to find the minimum value# of X using binary search.def findMin(a, b, n, k): # Boundaries of the Binary Search l = 0 u = max(a) while (l < u): # Find the middle value m = (l + u) // 2 # Check for the middle value if (check(a, b, k, n, m)): # Update the upper u = m else: # Update the lower l = m + 1 return l# Driver Codeif __name__ == '__main__': arr = [2, 1, 4, 3, 5] brr = [4, 3, 2, 3, 1] K = 12 N = len(arr) print(findMin(arr, brr, N, K)) # This code is contributed by ipg2016107. |
C#
// C# program for the above approach using System;public class GFG{ // Function to check if it is possible to// get desired resultstatic bool check(int []a, int []b, int k, int x){ int sum = 0; for(int i = 0; i < a.Length; i++) { sum = sum + (int)Math.Pow( Math.Max(a[i] - x, 0), b[i]); } if (sum <= k) return true; else return false;}// Function to find the minimum value // of X using binary search. static int findMin(int []a, int []b, int n, int k){ // Boundaries of the Binary Search int l = 0, u = (int)1e9; while (l < u) { // Find the middle value int m = (l + u) / 2; // Check for the middle value if (check(a, b, k, m)) // Update the upper u = m; else // Update the lower l = m + 1; } return l;}// Driver codepublic static void Main(String[] args){ int n = 5; int k = 12; int []a = { 2, 1, 4, 3, 5 }; int []b = { 4, 3, 2, 3, 1 }; Console.WriteLine(findMin(a, b, n, k));}}// This code is contributed by Princi Singh |
Javascript
<script> // JavaScript program for the above approache9 + 7; // Function to check if there exists an // X that satisfies the given conditions function check(a, b, k, n, x) { let sum = 0; // Find the required value of the // given expression for (let i = 0; i < n; i++) { sum = sum + Math.pow(Math.max(a[i] - x, 0), b[i]); } if (sum <= k) return true; else return false; } function max_element(a) { let maxi = Number.MIN_VALUE; for (let i = 0; i < a.length; i++) { if (a[i] > maxi) { maxi = a[i]; } } return maxi; } // Function to find the minimum value // of X using binary search. function findMin(a, b, n, k) { // Boundaries of the Binary Search let l = 0, u = max_element(a); while (l < u) { // Find the middle value let m = Math.floor((l + u) / 2); // Check for the middle value if (check(a, b, k, n, m)) { // Update the upper u = m; } else { // Update the lower l = m + 1; } } return l; } // Driver Code let arr = [2, 1, 4, 3, 5]; let brr = [4, 3, 2, 3, 1]; let K = 12; let N = arr.length; document.write(findMin(arr, brr, N, K));// This code is contributed by Potta Lokesh </script> |
Output
2
Time Complexity: O(N*log M), where, M is the maximum element of the array.
Auxiliary Space: O(1)
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