Find minimum speed to finish all Jobs
Given an array A and an integer H where and . Each element A[i] represents remaining pending jobs to be done and H represents hours left to complete all of the jobs. The task is to find the minimum speed in jobs per Hour at which person needs to work to complete all jobs in H hours.
Note: If A[i] has less job to be done than speed of person then he finish all of the work of A[i] and won’t go to next element during this hour.
Examples:
Input: A[] = [3, 6, 7, 11], H = 8 Output: 4 Input: A[] = [30, 11, 23, 4, 20], H = 5 Output: 30
Approach: If the person can finish all the jobs (within H hours) with an speed of K jobs/hour then he can finish all jobs with a larger speed too.
If we let ispossible(K) be true if and only if the person can finish with a job speed of K, then there is some X such that ispossible(K) = True if and only if K >= X.
For example, with A = [3, 6, 7, 11] and H = 8, there is some X = 4 so that ispossible(1) = ispossible(2) = ispossible(3) = False, and ispossible(4) = ispossible(5) = … = True.
We can do a binary search on the values of ispossible(K) to find the first X such that ispossible(X) is True which will be our answer.
Now, as it is not allowed to move from one element to other during the current hour even if the job is completed than the maximum possible value of K can be the maximum element in the array A[]. So, to find the value of ispossible(K), (i.e. whether person with a job speed of K can finish all jobs in H hours), do binary search in the range (1, max_element_of_array).
Also, for each A[i] of jobs > 0, we can deduce that person finishes it in Math.ceil(A[i] / K) or ((A[i]-1) / K) + 1 hours, and we add these for all elements to find the total time to complete all of the jobs and compare it with H to check if it is possible to finish all jobs with a speed of K jobs/hour.
Below is the implementation of above approach:
C++
// CPP program to find minimum speed // to finish all jobs #include <bits/stdc++.h> using namespace std; // Function to check if the person can do // all jobs in H hours with speed K bool isPossible( int A[], int n, int H, int K) { int time = 0; for ( int i = 0; i < n; ++i) time += (A[i] - 1) / K + 1; return time <= H; } // Function to return the minimum speed // of person to complete all jobs int minJobSpeed( int A[], int n, int H) { // If H < N it is not possible to complete // all jobs as person can not move from // one element to another during current hour if (H < n) return -1; // Max element of array int * max = max_element(A, A + n); int lo = 1, hi = *max; // Use binary search to find smallest K while (lo < hi) { int mi = lo + (hi - lo) / 2; if (!isPossible(A, n, H, mi)) lo = mi + 1; else hi = mi; } return lo; } // Driver program int main() { int A[] = { 3, 6, 7, 11 }, H = 8; int n = sizeof (A) / sizeof (A[0]); // Print required maxLenwer cout << minJobSpeed(A, n, H); return 0; } |
Java
// Java program to find minimum speed // to finish all jobs class GFG { // Function To findmax value in Array static int findmax( int [] A) { int r = A[ 0 ]; for ( int i = 1 ; i < A.length; i++) r = Math.max(r, A[i]); return r; } // Function to check if the person can do // all jobs in H hours with speed K static boolean isPossible( int [] A, int n, int H, int K) { int time = 0 ; for ( int i = 0 ; i < n; ++i) time += (A[i] - 1 ) / K + 1 ; return time <= H; } // Function to return the minimum speed // of person to complete all jobs static int minJobSpeed( int [] A, int n, int H) { // If H < N it is not possible to // complete all jobs as person can // not move from one element to // another during current hour if (H < n) return - 1 ; // Max element of array int max = findmax(A); int lo = 1 , hi = max; // Use binary search to find // smallest K while (lo < hi) { int mi = lo + (hi - lo) / 2 ; if (!isPossible(A, n, H, mi)) lo = mi + 1 ; else hi = mi; } return lo; } // Driver Code public static void main(String[] args) { int [] A = { 3 , 6 , 7 , 11 }; int H = 8 ; int n = A.length; // Print required maxLenwer System.out.println(minJobSpeed(A, n, H)); } } // This code is contributed by mits |
C#
// C# program to find minimum speed // to finish all jobs using System; using System.Linq; class GFG{ // Function to check if the person can do // all jobs in H hours with speed K static bool isPossible( int [] A, int n, int H, int K) { int time = 0; for ( int i = 0; i < n; ++i) time += (A[i] - 1) / K + 1; return time <= H; } // Function to return the minimum speed // of person to complete all jobs static int minJobSpeed( int [] A, int n, int H) { // If H < N it is not possible to complete // all jobs as person can not move from // one element to another during current hour if (H < n) return -1; // Max element of array int max = A.Max(); int lo = 1, hi = max; // Use binary search to find smallest K while (lo < hi) { int mi = lo + (hi - lo) / 2; if (!isPossible(A, n, H, mi)) lo = mi + 1; else hi = mi; } return lo; } // Driver program public static void Main() { int [] A = { 3, 6, 7, 11 }; int H = 8; int n = A.Length; // Print required maxLenwer Console.WriteLine(minJobSpeed(A, n, H)); } } |
Python3
# Python3 program to find minimum # speed to finish all jobs # Function to check if the person can do # all jobs in H hours with speed K def isPossible(A, n, H, K): time = 0 for i in range (n): time + = (A[i] - 1 ) / / K + 1 return time < = H # Function to return the minimum speed # of person to complete all jobs def minJobSpeed(A, n, H): # If H < N it is not possible to complete # all jobs as person can not move from # one element to another during current hour if H < n: return - 1 # Max element of array Max = max (A) lo, hi = 1 , Max # Use binary search to find smallest K while lo < hi: mi = lo + (hi - lo) / / 2 if not isPossible(A, n, H, mi): lo = mi + 1 else : hi = mi return lo if __name__ = = "__main__" : A = [ 3 , 6 , 7 , 11 ] H = 8 n = len (A) # Print required maxLenwer print (minJobSpeed(A, n, H)) # This code is contributed by Rituraj Jain |
Javascript
<script> // Javascript program to find minimum speed // to finish all jobs // Function to check if the person can do // all jobs in H hours with speed K function isPossible(A, n, H, K) { var time = 0; for ( var i = 0; i < n; ++i) time += parseInt((A[i] - 1) / K) + 1; return time <= H; } // Function to return the minimum speed // of person to complete all jobs function minJobSpeed(A, n, H) { // If H < N it is not possible to complete // all jobs as person can not move from // one element to another during current hour if (H < n) return -1; // Max element of array var max = A.reduce((a,b)=> Math.max(a,b)); var lo = 1, hi = max; // Use binary search to find smallest K while (lo < hi) { var mi = lo + parseInt((hi - lo) / 2); if (!isPossible(A, n, H, mi)) lo = mi + 1; else hi = mi; } return lo; } // Driver program var A = [3, 6, 7, 11], H = 8; var n = A.length; // Print required maxLenwer document.write( minJobSpeed(A, n, H)); // This code is contributed by famously. </script> |
Output:
4
Complexity analysis:
The time complexity of this program is O(N log M), where N is the length of the input array A and M is the maximum element in A. This is because the binary search is performed on the range [1, M], and for each mid-point, the isPossible function is called, which has a time complexity of O(N).
The space complexity of the program is O(1), as only a constant amount of memory is used to store variables that do not depend on the size of the input. The input array A is not modified during the execution of the program, so it does not contribute to the space complexity.