# Minimum time required to produce m items

• Difficulty Level : Medium
• Last Updated : 15 Dec, 2022

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

Method 1 : (Brute Force)
Initialize time = 0 and increment it by 1. Calculate the number of items produced at each time until the 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)``{``    ``// Initialize 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[0]);``    ``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)``{``    ` `    ``// Initialize 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):` `    ``# Initialize 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)``{``    ` `    ``// Initialize 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.``?>`

## Javascript

 ``

Output

```6
```

Time Complexity: O(n*m), it may vary for many test cases.
Space Complexity: O(1).

Method 2 (efficient):
The idea is to use Binary Search. The maximum possible time required to produce m items will be the 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 produces 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[0]);``    ``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.MIN_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 contributed 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.``?>`

## Javascript

 ``

Output

```6
```

Time Complexity : O(N * log M)
Auxiliary Space : O(1)

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