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 approach#include <bits/stdc++.h>using namespace std;// 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; // Perform the binary search while (low <= high) { // Find the value of mid int mid = (low + high) / 2; int sum = 0; // 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; // Check if temp is not // less than K if (temp >= K) { // Update the value // of sum sum += temp; } } // Check if the value of sum // is greater than X if (sum >= X) { // Update value of high minDays = mid; high = mid - 1; } // Update the value of low else { low = mid + 1; } } // Print the minimum number // of days cout << minDays;}// Driver Codeint main(){ int X = 100, K = 45; int arr[] = { 2, 5, 2, 6 }; int R[] = { 10, 13, 15, 12 }; int N = sizeof(arr) / sizeof(arr[0]); findMinDays(arr, R, N, X, K); return 0;} |
Java
// Java program for the above approachimport java.io.*;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; // Perform the binary search while (low <= high) { // Find the value of mid int mid = (low + high) / 2; int sum = 0; // 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; // Check if temp is not // less than K if (temp >= K) { // Update the value // of sum sum += temp; } } // Check if the value of sum // is greater than X if (sum >= X) { // Update value of high minDays = mid; high = mid - 1; } // Update the value of low else { low = mid + 1; } } // Print the minimum number // of days System.out.println(minDays);}// Driver Codepublic 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; findMinDays(arr, R, N, X, K);}}// This code is contributed by Potta Lokesh |
C#
// C# program for the above approachusing System;class GFG { // 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) { // Initialize the boundaries of // search space int low = 0, high = X; int minDays = -1; // Perform the binary search while (low <= high) { // Find the value of mid int mid = (low + high) / 2; int sum = 0; // 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; // Check if temp is not // less than K if (temp >= K) { // Update the value // of sum sum += temp; } } // Check if the value of sum // is greater than X if (sum >= X) { // Update value of high minDays = mid; high = mid - 1; } // Update the value of low else { low = mid + 1; } } // Print the minimum number // of days Console.Write(minDays); } // 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; findMinDays(arr, R, N, X, K); }}// This code is contributed by ukasp. |
Javascript
<script>// Javascript program for the above approach// 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; // Perform the binary search while (low <= high) { // Find the value of mid let mid = Math.floor((low + high) / 2); let sum = 0; // Traverse the array, arr[] for (let i = 0; i < N; i++) { // Find the value of arr[i] // after mid number of days let temp = arr[i] + R[i] * mid; // Check if temp is not // less than K if (temp >= K) { // Update the value // of sum sum += temp; } } // Check if the value of sum // is greater than X if (sum >= X) { // Update value of high minDays = mid; high = mid - 1; } // Update the value of low else { low = mid + 1; } } // Print the minimum number // of days document.write(minDays);}// 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);// This code is contributed by _saurabh_jaiswal.</script> |
Python3
# Python 3 program for the above approach# 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 = 0 # Perform the binary search while (low <= high): # Find the value of mid mid = (low + high) // 2 sum = 0 # 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] * mid # Check if temp is not # less than K if (temp >= K): # Update the value # of sum sum += temp # Check if the value of sum # is greater than X if (sum >= X): # Update value of high minDays = mid high = mid - 1 # Update the value of low else: low = mid + 1 # Print the minimum number # of days print(minDays)# Driver Codeif __name__ == '__main__': X = 100 K = 45 arr = [2, 5, 2, 6] R = [10, 13, 15, 12] N = len(arr) findMinDays(arr, R, N, X, K) # This code is contributed by SURENDRA_GANGWAR. |
Output:
4
Time Complexity: O(N*log X)
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!
