Given two integers X, K, and two arrays arr[] and R[] both consisting of N positive integers where R[i] denotes the amount by which arr[i] increases in one day, the task is to find the minimum number of days after which the sum of array elements having value greater than or equal to K becomes at least X.
Examples:
Input: X = 100, K = 45, arr[] = {2, 5, 2, 6}, R[] = {10, 13, 15, 12}
Output: 4
Explanation:
Consider the following values of array after each day:
- Day 1: After the day 1, all array element modifies to {12, 18, 17, 18}. The sum of elements having values >= K(= 45) is 0.
- Day 2: After the day 2, all array element modifies to {22, 31, 32, 30}. The sum of elements having values >= K(= 45) is 0.
- Day 3: After the day 3, all array element modifies to {32, 44, 47, 42}. The sum of elements having values >= K(= 45) is 47.
- Day 4: After the day 4, all array element modifies to {42, 57, 62, 54}. The sum of elements having values >= K(= 45) is 57 + 62 + 54 = 167, which is at least X(= 100).
Therefore, the minimum number of days required is 4.
Input: X = 65, K = 10, arr[] = {1, 1, 1, 1, 3}, R[] = {2, 1, 2, 2, 1}
Output: 9
Naive Approach: The simplest approach to solve the given problem is to keep incrementing the number of days and whenever the sum of the array elements having a value at least K becomes greater than or equal to X. After incrementing for D days, print the value of the current number of days obtained.
Time Complexity: O(N*X)
Auxiliary Space: O(1)
Efficient Approach: The above approach can also be optimized by using Binary Search. Follow the steps below to solve the problem:
- Initialize two variables, say low as 0 and high as X.
- Initialize a variable, say minDays that stores the minimum number of days.
- Iterate until the value of low is at most high and perform the following steps:
- Initialize a variable mid as low + (high – low)/2 and variable, say sum as 0 to store the sum of array elements after mid number of days.
- Traverse the array, arr[] using the variable i and perform the following steps:
- Initialize a variable temp as (arr[i] + R[i]*mid).
- If the value of temp is not less than K add the value of temp to sum.
- If the value of sum is at least X, then update the value of minDays to mid and the value of high to (mid – 1).
- Otherwise, update the value of low to (mid + 1).
- After completing the above steps, print the value of minDays as the resultant minimum number of days.
Below is the implementation of the above approach:
C++
// C++ program for the above approachÂ
#include <bits/stdc++.h>using namespace std;Â
// Function to find the minimum number// of days such that the sum of array// elements >= K is at least Xvoid findMinDays(int arr[], int R[],                 int N, int X, int K){    // Initialize the boundaries of    // search space    int low = 0, high = X;    int minDays;Â
    // Perform the binary search    while (low <= high) {Â
        // Find the value of mid        int mid = (low + high) / 2;Â
        int sum = 0;Â
        // Traverse the array, arr[]        for (int i = 0; i < N; i++) {Â
            // Find the value of arr[i]            // after mid number of days            int temp = arr[i] + R[i] * mid;Â
            // Check if temp is not            // less than K            if (temp >= K) {Â
                // Update the value                // of sum                sum += temp;            }        }Â
        // Check if the value of sum        // is greater than X        if (sum >= X) {Â
            // Update value of high            minDays = mid;            high = mid - 1;        }Â
        // Update the value of low        else {            low = mid + 1;        }    }Â
    // Print the minimum number    // of days    cout << minDays;}Â
// Driver Codeint main(){Â Â Â Â int X = 100, K = 45;Â Â Â Â int arr[] = { 2, 5, 2, 6 };Â Â Â Â int R[] = { 10, 13, 15, 12 };Â Â Â Â int N = sizeof(arr) / sizeof(arr[0]);Â Â Â Â findMinDays(arr, R, N, X, K);Â
    return 0;} |
Java
// Java program for the above approachimport java.io.*;Â
class GFG{     // Function to find the minimum number// of days such that the sum of array// elements >= K is at least Xstatic void findMinDays(int arr[], int R[], int N,                        int X, int K){         // Initialize the boundaries of    // search space    int low = 0, high = X;    int minDays = -1;Â
    // Perform the binary search    while (low <= high)     {                 // Find the value of mid        int mid = (low + high) / 2;Â
        int sum = 0;Â
        // Traverse the array, arr[]        for(int i = 0; i < N; i++)        {                         // Find the value of arr[i]            // after mid number of days            int temp = arr[i] + R[i] * mid;Â
            // Check if temp is not            // less than K            if (temp >= K)             {Â
                // Update the value                // of sum                sum += temp;            }        }Â
        // Check if the value of sum        // is greater than X        if (sum >= X)         {                         // Update value of high            minDays = mid;            high = mid - 1;        }Â
        // Update the value of low        else        {            low = mid + 1;        }    }Â
    // Print the minimum number    // of days    System.out.println(minDays);}Â
// Driver Codepublic static void main(String[] args){Â Â Â Â int X = 100, K = 45;Â Â Â Â int arr[] = { 2, 5, 2, 6 };Â Â Â Â int R[] = { 10, 13, 15, 12 };Â Â Â Â int N = arr.length;Â Â Â Â Â Â Â Â Â findMinDays(arr, R, N, X, K);}}Â
// This code is contributed by Potta Lokesh |
C#
// C# program for the above approachusing System;class GFG {Â
    // Function to find the minimum number    // of days such that the sum of array    // elements >= K is at least X    static void findMinDays(int[] arr, int[] R, int N,                            int X, int K)    {Â
        // Initialize the boundaries of        // search space        int low = 0, high = X;        int minDays = -1;Â
        // Perform the binary search        while (low <= high) {Â
            // Find the value of mid            int mid = (low + high) / 2;Â
            int sum = 0;Â
            // Traverse the array, arr[]            for (int i = 0; i < N; i++) {Â
                // Find the value of arr[i]                // after mid number of days                int temp = arr[i] + R[i] * mid;Â
                // Check if temp is not                // less than K                if (temp >= K) {Â
                    // Update the value                    // of sum                    sum += temp;                }            }Â
            // Check if the value of sum            // is greater than X            if (sum >= X) {Â
                // Update value of high                minDays = mid;                high = mid - 1;            }Â
            // Update the value of low            else {                low = mid + 1;            }        }Â
        // Print the minimum number        // of days        Console.Write(minDays);    }Â
    // Driver Code    public static void Main(string[] args)    {        int X = 100, K = 45;        int[] arr = { 2, 5, 2, 6 };        int[] R = { 10, 13, 15, 12 };        int N = arr.Length;Â
        findMinDays(arr, R, N, X, K);    }}Â
// This code is contributed by ukasp. |
Javascript
<script>Â
// Javascript program for the above approachÂ
Â
// Function to find the minimum number// of days such that the sum of array// elements >= K is at least Xfunction findMinDays(arr, R, N, X, K) {    // Initialize the boundaries of    // search space    let low = 0, high = X;    let minDays;Â
    // Perform the binary search    while (low <= high) {Â
        // Find the value of mid        let mid = Math.floor((low + high) / 2);Â
        let sum = 0;Â
        // Traverse the array, arr[]        for (let i = 0; i < N; i++) {Â
            // Find the value of arr[i]            // after mid number of days            let temp = arr[i] + R[i] * mid;Â
            // Check if temp is not            // less than K            if (temp >= K) {Â
                // Update the value                // of sum                sum += temp;            }        }Â
        // Check if the value of sum        // is greater than X        if (sum >= X) {Â
            // Update value of high            minDays = mid;            high = mid - 1;        }Â
        // Update the value of low        else {            low = mid + 1;        }    }Â
    // Print the minimum number    // of days    document.write(minDays);}Â
// Driver Codelet X = 100, K = 45;let arr = [2, 5, 2, 6];let R = [10, 13, 15, 12];let N = arr.lengthfindMinDays(arr, R, N, X, K);Â
// This code is contributed by _saurabh_jaiswal.</script> |
Python3
# Python 3 program for the above approachÂ
# Function to find the minimum number# of days such that the sum of array# elements >= K is at least Xdef findMinDays(arr, R, N, X, K):       # Initialize the boundaries of    # search space    low = 0    high = X    minDays = 0Â
    # Perform the binary search    while (low <= high):        # Find the value of mid        mid = (low + high) // 2Â
        sum = 0Â
        # Traverse the array, arr[]        for i in range(N):            # Find the value of arr[i]            # after mid number of days            temp = arr[i] + R[i] * midÂ
            # Check if temp is not            # less than K            if (temp >= K):                # Update the value                # of sum                sum += tempÂ
        # Check if the value of sum        # is greater than X        if (sum >= X):Â
            # Update value of high            minDays = mid            high = mid - 1Â
        # Update the value of low        else:            low = mid + 1Â
    # Print the minimum number    # of days    print(minDays)Â
# Driver Codeif __name__ == '__main__':Â Â Â Â X = 100Â Â Â Â K = 45Â Â Â Â arr = [2, 5, 2, 6]Â Â Â Â R = [10, 13, 15, 12]Â Â Â Â N = len(arr)Â Â Â Â findMinDays(arr, R, N, X, K)Â Â Â Â Â Â Â Â Â # This code is contributed by SURENDRA_GANGWAR. |
Output:Â
4
Â
Time Complexity: O(N*log X)
Auxiliary Space: O(1)
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