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Minimum value of X such that sum of arr[i] – X raised to the power of brr[i] is less than or equal to K

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  • Last Updated : 09 Aug, 2021
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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 conditions
bool 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 Code
int 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 result
static 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 code
public 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 conditions
def 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 Code
if __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 result
static 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 code
public 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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