# 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; ` `} ` |

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## 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 ` |

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## 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)); ` `} ` `} ` |

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## 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 [tabbyending]

**Output:**

4

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

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