Given an array arr[] consisting of N integers representing the number of coins in each pile, and an integer H, the task is to find the minimum number of coins that must be collected from a single pile per hour such that all the piles are emptied in less than H hours.
Note: Coins can be collected only from a single pile in an hour.
Examples:
Input: arr[] = {3, 6, 7, 11}, H = 8
Output: 4
Explanation:
Removing 4 coins per pile in each hour, the time taken to empty each pile are as follows:
arr[0] = 3: Emptied in 1 hour.
arr[1] = 6: 4 coins removed in the 1st hour and 2 removed in the 2nd hour. Therefore, emptied in 2 hours.
arr[2] = 7: 4 coins removed in the 1st hour and 3 removed in the 2nd hour. Therefore, emptied in 2 hours.
arr[3] = 11: 4 coins removed in both 1st and 2nd hour, and 3 removed in the 3rd hour. Therefore, emptied in 3 hours.
Therefore, number of hours required = 1 + 2 + 2 + 3 = 8 ( = H).Input: arr[] = {30, 11, 23, 4, 20}, H = 5
Output: 30
Approach: The idea is to use Binary Search. Follow the steps below to solve the problem:
- Initialize a variable, say ans, to store the minimum number coins required to be collected per hour.
- Initialize variables low and high, as 1 and the maximum value present in the array, to set the range to perform Binary Search.
- Iterate until low ? high and perform the following steps:
- Find the value of mid as (low + high)/2.
- Traverse the array arr[] to find the time taken to empty all the pile of coins by removing mid coins per hour and check if the total time exceeds H or not. If found to be false, update the high to (K – 1) and update ans to K. Otherwise, update low to (K + 1).
- After completing the above steps, print the value of ans as the result.
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 coins to be collected per hour// to empty N piles in H hoursint minCollectingSpeed(vector<int>& piles, int H){ // Stores the minimum coins // to be removed per hour int ans = -1; int low = 1, high; // Find the maximum array element high = *max_element(piles.begin(), piles.end()); // Perform Binary Search while (low <= high) { // Store the mid value of the // range in K int K = low + (high - low) / 2; int time = 0; // Find the total time taken to // empty N piles by removing K // coins per hour for (int ai : piles) { time += (ai + K - 1) / K; } // If total time does not exceed H if (time <= H) { ans = K; high = K - 1; } // Otherwise else { low = K + 1; } } // Print the required result cout << ans;}// Driver Codeint main(){ vector<int> arr = { 3, 6, 7, 11 }; int H = 8; // Function Call minCollectingSpeed(arr, H); return 0;} |
Java
// Java program for the above approachimport java.util.*; class GFG{ // Function to find the minimum number// of coins to be collected per hour// to empty N piles in H hoursstatic void minCollectingSpeed(int[] piles, int H){ // Stores the minimum coins // to be removed per hour int ans = -1; int low = 1, high; // Find the maximum array element high = Arrays.stream(piles).max().getAsInt(); // Perform Binary Search while (low <= high) { // Store the mid value of the // range in K int K = low + (high - low) / 2; int time = 0; // Find the total time taken to // empty N piles by removing K // coins per hour for(int ai : piles) { time += (ai + K - 1) / K; } // If total time does not exceed H if (time <= H) { ans = K; high = K - 1; } // Otherwise else { low = K + 1; } } // Print the required result System.out.print(ans);} // Driver Codestatic public void main(String args[]){ int[] arr = { 3, 6, 7, 11 }; int H = 8; // Function Call minCollectingSpeed(arr, H);}}// This code is contributed by sanjoy_62 |
C#
// C# program for the above approachusing System;using System.Collections;class GFG{ // Function to find the minimum number // of coins to be collected per hour // to empty N piles in H hours static void minCollectingSpeed(int[] piles, int H) { // Stores the minimum coins // to be removed per hour int ans = -1; int low = 1, high; Array.Sort(piles); // Find the maximum array element high = piles[piles.Length - 1 ]; // Perform Binary Search while (low <= high) { // Store the mid value of the // range in K int K = low + (high - low) / 2; int time = 0; // Find the total time taken to // empty N piles by removing K // coins per hour foreach(int ai in piles) { time += (ai + K - 1) / K; } // If total time does not exceed H if (time <= H) { ans = K; high = K - 1; } // Otherwise else { low = K + 1; } } // Print the required result Console.Write(ans); } // Driver Code static public void Main(string []args) { int[] arr = { 3, 6, 7, 11 }; int H = 8; // Function Call minCollectingSpeed(arr, H); }}// This code is contributed by AnkThon |
Python3
# Python3 program for the above approach# Function to find the minimum number# of coins to be collected per hour# to empty N piles in H hoursdef minCollectingSpeed(piles, H): # Stores the minimum coins # to be removed per hour ans = -1 low = 1 # Find the maximum array element high = max(piles) # Perform Binary Search while (low <= high): # Store the mid value of the # range in K K = low + (high - low) // 2 time = 0 # Find the total time taken to # empty N piles by removing K # coins per hour for ai in piles: time += (ai + K - 1) // K # If total time does not exceed H if (time <= H): ans = K high = K - 1 # Otherwise else: low = K + 1 # Print required result print(ans)# Driver Codeif __name__ == '__main__': arr = [3, 6, 7, 11] H = 8 # Function Call minCollectingSpeed(arr, H)# This code is contributed by mohit kumar 29 |
Javascript
<script>// Javascript program for the above approach// Function to find the minimum number// of coins to be collected per hour// to empty N piles in H hoursfunction minCollectingSpeed(piles, H){ // Stores the minimum coins // to be removed per hour var ans = -1; var low = 1, high; // Find the maximum array element high = piles.reduce((a,b)=> Math.max(a,b)); // Perform Binary Search while (low <= high) { // Store the mid value of the // range in K var K = low + parseInt((high - low) / 2); var time = 0; // Find the total time taken to // empty N piles by removing K // coins per hour piles.forEach(ai => { time += parseInt((ai + K - 1) / K); }); // If total time does not exceed H if (time <= H) { ans = K; high = K - 1; } // Otherwise else { low = K + 1; } } // Print the required result document.write( ans);}// Driver Codevar arr = [3, 6, 7, 11];var H = 8;// Function CallminCollectingSpeed(arr, H);</script> |
Output:
4
Time Complexity: O(H*log N) // time complexity of binary search algorithm is logarithmic so the algorithm runs in logarithmic time
Auxiliary Space: O(1) // since no extra array is used hence space required by the algorithm is constant
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