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Find minimum speed to finish all Jobs

  • Difficulty Level : Medium
  • Last Updated : 26 May, 2021

Given an array A and an integer H where 1\leq A[i] \leq 10^{9}  and len(A) \leq H \leq 10^{9}  . 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

 

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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

 

Time Complexity: O(N*log(M)), where N is the length of array and M is max(A).
 




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