# Minimum time required to produce m items

Consider n machines which produce same type of items but at different rate i.e., machine 1 takes a1 sec to produce an item, machine 2 takes a2 sec to produce an item. Given an array which contains the time required by ith machine to produce an item. Considering all machine are working simultaneously, the task is to find minimum time required to produce m items.

Examples:

```Input : arr[] = {1, 2, 3}, m = 11
Output : 6
In 6 sec, machine 1 produces 6 items, machine 2 produces 3 items,
and machine 3 produces 2 items. So to produce 11 items minimum
6 sec are required.

Input : arr[] = {5, 6}, m = 11
Output : 30
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Method 1 : (Brute Force)
Initialize time = 0 and increment it by 1. Calculate number of item produce at each time until number of produced items is not equal to m.
Below is the implementation of the above idea :

## C++

 `// C++ program to find minimum time ` `// required to produce m items. ` `#include ` `using` `namespace` `std; ` ` `  `// Return the minimum time required to ` `// produce m items with given machines. ` `int` `minTime(``int` `arr[], ``int` `n, ``int` `m) ` `{ ` `    ``// Intialise time, items equal to 0. ` `    ``int` `t = 0; ` ` `  `    ``while` `(1) ` `    ``{ ` `        ``int` `items = 0; ` ` `  `        ``// Calculating items at each second ` `        ``for` `(``int` `i = 0; i < n; i++) ` `            ``items += (t / arr[i]); ` ` `  `        ``// If items equal to m return time. ` `        ``if` `(items >= m) ` `            ``return` `t; ` ` `  `        ``t++; ``// Increment time ` `    ``} ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 2, 3 }; ` `    ``int` `n = ``sizeof``(arr)/``sizeof``(arr); ` `    ``int` `m = 11; ` ` `  `    ``cout << minTime(arr, n, m) << endl; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find minimum time ` `// required to produce m items. ` `import` `java.io.*; ` ` `  `public` `class` `GFG{ ` ` `  `// Return the minimum time required to ` `// produce m items with given machines. ` `static` `int` `minTime(``int` `[]arr, ``int` `n,  ` `                   ``int` `m) ` `{ ` `     `  `    ``// Intialise time, items equal to 0. ` `    ``int` `t = ``0``; ` ` `  `    ``while` `(``true``) ` `    ``{ ` `        ``int` `items = ``0``; ` ` `  `        ``// Calculating items at each second ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `            ``items += (t / arr[i]); ` ` `  `        ``// If items equal to m return time. ` `        ``if` `(items >= m) ` `            ``return` `t; ` ` `  `        ``t++; ``// Increment time ` `    ``} ` `} ` ` `  `    ``// Driver Code ` `    ``static` `public` `void` `main (String[] args) ` `    ``{ ` `            ``int` `[]arr = { ``1``, ``2``, ``3` `}; ` `            ``int` `n = arr.length; ` `            ``int` `m = ``11``; ` ` `  `    ``System.out.println(minTime(arr, n, m)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## Python3

 `# Python3 program to find minimum time ` `# required to produce m items. ` `import` `math as mt ` ` `  `# Return the minimum time required to ` `# produce m items with given machines. ` `def` `minTime(arr, n, m): ` ` `  `    ``# Intialise time, items equal to 0. ` `    ``t ``=` `0` ` `  `    ``while` `(``1``): ` `     `  `        ``items ``=` `0` ` `  `        ``# Calculating items at each second ` `        ``for` `i ``in` `range``(n): ` `            ``items ``+``=` `(t ``/``/` `arr[i]) ` ` `  `        ``# If items equal to m return time. ` `        ``if` `(items >``=` `m): ` `            ``return` `t ` ` `  `        ``t ``+``=` `1` `# Increment time ` `     `  `# Driver Code ` `arr ``=` `[``1``, ``2``, ``3``]  ` `n ``=` `len``(arr) ` `m ``=` `11` ` `  `print``(minTime(arr, n, m) ) ` ` `  `# This code is contributed by  ` `# Mohit kumar 29 `

## C#

 `// C# program to find minimum time ` `// required to produce m items. ` `using` `System; ` `  `  `public` `class` `GFG{ ` `  `  `// Return the minimum time  ` `// required to produce m  ` `// items with given machines. ` `static` `int` `minTime(``int` `[]arr, ``int` `n,  ` `                   ``int` `m) ` `{ ` `     `  `    ``// Intialise time, items equal to 0. ` `    ``int` `t = 0; ` `  `  `    ``while` `(``true``) ` `    ``{ ` `        ``int` `items = 0; ` `  `  `        ``// Calculating items at each second ` `        ``for` `(``int` `i = 0; i < n; i++) ` `            ``items += (t / arr[i]); ` `  `  `        ``// If items equal to m return time. ` `        ``if` `(items >= m) ` `            ``return` `t; ` `  `  `        ``t++; ``// Increment time ` `    ``} ` `} ` `  `  `// Driven Code ` `static` `public` `void` `Main (String []args) ` `{ ` `    ``int` `[]arr = {1, 2, 3}; ` `    ``int` `n = arr.Length; ` `    ``int` `m = 11; ` `  `  `    ``// Calling function ` `    ``Console.WriteLine(minTime(arr, n, m)); ` `     `  `} ` `} `

## PHP

 `= ``\$m``) ` `            ``return` `\$t``; ` ` `  `        ``\$t``++; ``// Increment time ` `    ``} ` `} ` ` `  `    ``// Driver Code ` `    ``\$arr` `= ``array``(1, 2, 3); ` `    ``\$n` `=``count``(``\$arr``); ` `    ``\$m` `= 11; ` `    ``echo` `minTime(``\$arr``, ``\$n``, ``\$m``); ` ` `  `// This code is contributed by anuj_67. ` `?> `

Output:

```6
```

Method 2 (efficient):
The idea is to use Binary Search. Maximum possible time required to produce m items will be maximum time in the array, amax, multiplied by m i.e amax * m. So, use binary search between 1 to amax * m and find the minimum time which produce m items.
Below is the implementation of the above idea :

## C++

 `// Efficient C++ program to find minimum time ` `// required to produce m items. ` `#include ` `using` `namespace` `std; ` ` `  `// Return the number of items can be ` `// produced in temp sec. ` `int` `findItems(``int` `arr[], ``int` `n, ``int` `temp) ` `{ ` `    ``int` `ans = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``ans += (temp/arr[i]); ` `    ``return` `ans; ` `} ` ` `  `// Binary search to find minimum time required ` `// to produce M items. ` `int` `bs(``int` `arr[], ``int` `n, ``int` `m, ``int` `high) ` `{ ` `    ``int` `low = 1; ` ` `  `    ``// Doing binary search to find minimum ` `    ``// time. ` `    ``while` `(low < high) ` `    ``{ ` `        ``// Finding the middle value. ` `        ``int` `mid = (low+high)>>1; ` ` `  `        ``// Calculate number of items to ` `        ``// be produce in mid sec. ` `        ``int` `itm = findItems(arr, n, mid); ` ` `  `        ``// If items produce is less than ` `        ``// required, set low = mid + 1. ` `        ``if` `(itm < m) ` `            ``low = mid+1; ` ` `  `        ``//  Else set high = mid. ` `        ``else` `            ``high = mid; ` `    ``} ` ` `  `    ``return` `high; ` `} ` ` `  `// Return the minimum time required to ` `// produce m items with given machine. ` `int` `minTime(``int` `arr[], ``int` `n, ``int` `m) ` `{ ` `    ``int` `maxval = INT_MIN; ` ` `  `    ``// Finding the maximum time in the array. ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``maxval = max(maxval, arr[i]); ` ` `  `    ``return` `bs(arr, n, m, maxval*m); ` `} ` ` `  `// Driven Program ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 2, 3 }; ` `    ``int` `n = ``sizeof``(arr)/``sizeof``(arr); ` `    ``int` `m = 11; ` ` `  `    ``cout << minTime(arr, n, m) << endl; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Efficient Java program to find  ` `// minimum time required to  ` `// produce m items. ` `import` `java.io.*; ` ` `  `public` `class` `GFG{ ` `     `  `// Return the number of items can  ` `// be produced in temp sec. ` `static` `int` `findItems(``int` `[]arr, ``int` `n,  ` `                     ``int` `temp) ` `{ ` `    ``int` `ans = ``0``; ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``ans += (temp / arr[i]); ` `    ``return` `ans; ` `} ` ` `  `// Binary search to find minimum time  ` `// required to produce M items. ` `static` `int` `bs(``int` `[]arr, ``int` `n,  ` `              ``int` `m, ``int` `high) ` `              `  `{ ` `    ``int` `low = ``1``; ` ` `  `    ``// Doing binary search to  ` `    ``// find minimum time. ` `    ``while` `(low < high) ` `    ``{ ` `        ``// Finding the middle value. ` `        ``int` `mid = (low + high) >> ``1``; ` ` `  `        ``// Calculate number of items to ` `        ``// be produce in mid sec. ` `        ``int` `itm = findItems(arr, n, mid); ` ` `  `        ``// If items produce is less than ` `        ``// required, set low = mid + 1. ` `        ``if` `(itm < m) ` `            ``low = mid + ``1``; ` ` `  `        ``// Else set high = mid. ` `        ``else` `            ``high = mid; ` `    ``} ` ` `  `    ``return` `high; ` `} ` ` `  `// Return the minimum time required to ` `// produce m items with given machine. ` `static` `int` `minTime(``int` `[]arr, ``int` `n,  ` `                   ``int` `m) ` `{ ` `    ``int` `maxval = Integer.MAX_VALUE; ` ` `  `    ``// Finding the maximum time in the array. ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``maxval = Math.max(maxval, arr[i]); ` ` `  `    ``return` `bs(arr, n, m, maxval * m); ` `} ` ` `  `// Driven Program ` `static` `public` `void` `main (String[] args) ` `{ ` `    ``int` `[]arr = {``1``, ``2``, ``3``}; ` `        ``int` `n = arr.length; ` `    ``int` `m = ``11``; ` ` `  `    ``System.out.println(minTime(arr, n, m)); ` `} ` `} ` ` `  `// This code is contributed by vt_m. `

## Python3

 `# Efficient Python3 program to find  ` `# minimum time required to produce m items. ` `import` `sys ` ` `  `def` `findItems(arr, n, temp): ` `    ``ans ``=` `0` `    ``for` `i ``in` `range``(n): ` `        ``ans ``+``=` `temp ``/``/` `arr[i] ` `    ``return` `ans ` ` `  `# Binary search to find minimum time  ` `# required to produce M items. ` `def` `bs(arr, n, m, high): ` `    ``low ``=` `1` ` `  `    ``# Doing binary search to find minimum ` `    ``# time. ` `    ``while` `low < high: ` ` `  `        ``# Finding the middle value. ` `        ``mid ``=` `(low ``+` `high) >> ``1` ` `  `        ``# Calculate number of items to ` `        ``# be produce in mid sec. ` `        ``itm ``=` `findItems(arr, n, mid) ` ` `  `        ``# If items produce is less than ` `        ``# required, set low = mid + 1. ` `        ``if` `itm < m: ` `            ``low ``=` `mid ``+` `1` ` `  `        ``# Else set high = mid. ` `        ``else``: ` `            ``high ``=` `mid ` `    ``return` `high ` ` `  `# Return the minimum time required to ` `# produce m items with given machine. ` `def` `minTime(arr, n, m): ` `    ``maxval ``=` `-``sys.maxsize ` ` `  `    ``# Finding the maximum time in the array. ` `    ``for` `i ``in` `range``(n): ` `        ``maxval ``=` `max``(maxval, arr[i]) ` ` `  `    ``return` `bs(arr, n, m, maxval ``*` `m) ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` `    ``arr ``=` `[``1``, ``2``, ``3``] ` `    ``n ``=` `len``(arr) ` `    ``m ``=` `11` `    ``print``(minTime(arr, n, m)) ` ` `  `# This code is conributed by ` `# sanjeev2552 `

## C#

 `// Efficient C# program to find  ` `// minimum time required to  ` `// produce m items. ` `using` `System; ` ` `  `public` `class` `GFG{ ` `     `  `// Return the number of items can  ` `// be produced in temp sec. ` `static` `int` `findItems(``int` `[]arr, ``int` `n,  ` `                     ``int` `temp) ` `{ ` `    ``int` `ans = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``ans += (temp / arr[i]); ` `    ``return` `ans; ` `} ` ` `  `// Binary search to find minimum time  ` `// required to produce M items.. ` `static` `int` `bs(``int` `[]arr, ``int` `n,  ` `              ``int` `m, ``int` `high) ` `{ ` `    ``int` `low = 1; ` ` `  `    ``// Doing binary search to  ` `    ``// find minimum time. ` `    ``while` `(low < high) ` `    ``{ ` `        ``// Finding the middle value. ` `        ``int` `mid = (low + high) >> 1; ` ` `  `        ``// Calculate number of items to ` `        ``// be produce in mid sec. ` `        ``int` `itm = findItems(arr, n, mid); ` ` `  `        ``// If items produce is less than ` `        ``// required, set low = mid + 1. ` `        ``if` `(itm < m) ` `            ``low = mid + 1; ` ` `  `        ``// Else set high = mid. ` `        ``else` `            ``high = mid; ` `    ``} ` ` `  `    ``return` `high; ` `} ` ` `  `// Return the minimum time required to ` `// produce m items with given machine. ` `static` `int` `minTime(``int` `[]arr, ``int` `n,  ` `                   ``int` `m) ` `{ ` `    ``int` `maxval = ``int``.MinValue; ` ` `  `    ``// Finding the maximum time in the array. ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``maxval = Math.Max(maxval, arr[i]); ` ` `  `    ``return` `bs(arr, n, m, maxval * m); ` `} ` ` `  `// Driver Code ` `static` `public` `void` `Main () ` `{ ` `    ``int` `[]arr = {1, 2, 3}; ` `    ``int` `n = arr.Length; ` `    ``int` `m = 11; ` ` `  `    ``Console.WriteLine(minTime(arr, n, m)); ` `} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 `> 1; ` ` `  `        ``// Calculate number of items to ` `        ``// be produce in mid sec. ` `        ``\$itm` `= findItems(``\$arr``, ``\$n``, ``\$mid``); ` ` `  `        ``// If items produce is less than ` `        ``// required, set low = mid + 1. ` `        ``if` `(``\$itm` `< ``\$m``) ` `            ``\$low` `= ``\$mid` `+ 1; ` ` `  `        ``// Else set high = mid. ` `        ``else` `            ``\$high` `= ``\$mid``; ` `    ``} ` ` `  `    ``return` `\$high``; ` `} ` ` `  `// Return the minimum time required to ` `// produce m items with given machine. ` `function` `minTime(``\$arr``, ``\$n``, ``\$m``) ` `{ ` `    ``\$maxval` `= PHP_INT_MIN; ` ` `  `    ``// Finding the maximum ` `    ``// time in the array. ` `    ``for``(``\$i` `= 0; ``\$i` `< ``\$n``; ``\$i``++) ` `        ``\$maxval` `= max(``\$maxval``, ``\$arr``[``\$i``]); ` ` `  `    ``return` `bs(``\$arr``, ``\$n``, ``\$m``, ``\$maxval``*``\$m``); ` `} ` ` `  `    ``// Driver Code ` `    ``\$arr` `= ``array``(1, 2, 3); ` `    ``\$n` `= ``count``(``\$arr``); ` `    ``\$m` `= 11; ` ` `  `    ``echo` `minTime(``\$arr``, ``\$n``, ``\$m``); ` ` `  `// This code is contributed by anuj_67. ` `?> `

Output:

```6
```

This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.