Minimum days to make Array elements with value at least K sum at least X

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:

1. 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.
2. 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.
3. 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.
4. 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++

#include <bits/stdc++.h> using namespace std;    void findMinDays(int arr[], int R[],                  int N, int X, int K) {               int low = 0, high = X;     int minDays;             while (low <= high) {                     int mid = (low + high) / 2;            int sum = 0;                     for (int i = 0; i < N; i++) {                                          int temp = arr[i] + R[i] * mid;                                          if (temp >= K) {                                                      sum += temp;             }         }                              if (sum >= X) {                             minDays = mid;             high = mid - 1;         }                     else {             low = mid + 1;         }     }                  cout << minDays; }    int 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; }

Time Complexity: O(N*log X)
Auxiliary Space: O(1)

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