Given an array of positive distinct integer denoting the crossing time of ‘n’ people. These ‘n’ people are standing at one side of bridge. Bridge can hold at max two people at a time. When two people cross the bridge, they must move at the slower person’s pace. Find the minimum total time in which all persons can cross the bridge. *See this puzzle to understand more*.**Note:** Slower person’pace is given by larger time.

Input:Crossing Times = {10, 20, 30}Output:60Explanation1. Firstly person '1' and '2' cross the bridge with total time about 20 min(maximum of 10, 20) 2. Now the person '1' will come back with total time of '10' minutes. 3. Lastly the person '1' and '3' cross the bridge with total time about 30 minutes Hence total time incurred in whole journey will be 20 + 10 + 30 = 60Input:Crossing Times = [1, 2, 5, 8}Output:15ExplanationSee this for full explanation.

The approach is to use Dynamic programming. Before getting dive into dynamic programminc let’s see the following observation that will be required in solving the problem.

- When any two people cross the bridge, then the fastest person crossing time will not be contributed in answer as both of them move with slowest person speed.
- When some of the people will cross the river and reached the right side then only the fastest people(smallest integer) will come back to the left side.
- Person can only be present either left side or right side of the bridge. Thus, if we maintain the left mask, then right mask can easily be calculated by setting the bits ‘1’ which is not present in the left mask. For instance, Right_mask = ((2
^{n}) – 1) XOR (left_mask). - Any person can easily be represented by bitmask(usually called as ‘mask’). When i
^{th}bit of ‘mask’ is set, that means that person is present at left side of the bridge otherwise it would be present at right side of bridge. For instance, let the mask of**6 people**is 100101, which reprsents the person 1, 4, 6 are present at left side of bridge and the person 2, 3 and 5 are present at the right side of the bridge.

`// C++ program to find minimum time required to` `// send people on other side of bridge` `#include <bits/stdc++.h>` `using` `namespace` `std;` ` ` `/* Global dp[2^20][2] array, in dp[i][j]--` ` ` `'i' denotes mask in which 'set bits' denotes` ` ` `total people standing at left side of bridge` ` ` `and 'j' denotes the turn that represent on ` ` ` `which side we have to send people either` ` ` `from left to right(0) or from right to ` ` ` `left(1) */` `int` `dp[1 << 20][2];` ` ` `/* Utility function to find total time required` ` ` `to send people to other side of bridge */` `int` `findMinTime(` `int` `leftmask, ` `bool` `turn, ` `int` `arr[], ` `int` `& n)` `{` ` ` ` ` `// If all people has been transfered` ` ` `if` `(!leftmask)` ` ` `return` `0;` ` ` ` ` `int` `& res = dp[leftmask][turn];` ` ` ` ` `// If we already have solved this subproblem, ` ` ` `// return the answer.` ` ` `if` `(~res)` ` ` `return` `res;` ` ` ` ` `// Calculate mask of right side of people` ` ` `int` `rightmask = ((1 << n) - 1) ^ leftmask;` ` ` ` ` `/* if turn == 1 means currently people are at` ` ` `right side, thus we need to transfer` ` ` `people to the left side */` ` ` `if` `(turn == 1) {` ` ` `int` `minRow = INT_MAX, person;` ` ` `for` `(` `int` `i = 0; i < n; ++i) {` ` ` ` ` `// Select one people whose time is less` ` ` `// among all others present at right` ` ` `// side` ` ` `if` `(rightmask & (1 << i)) {` ` ` `if` `(minRow > arr[i]) {` ` ` `person = i;` ` ` `minRow = arr[i];` ` ` `}` ` ` `}` ` ` `}` ` ` ` ` `// Add that person to answer and recurse for next turn` ` ` `// after initializing that person at left side` ` ` `res = arr[person] + findMinTime(leftmask | (1 << person),` ` ` `turn ^ 1, arr, n);` ` ` `}` ` ` `else` `{` ` ` ` ` `// __builtin_popcount() is inbuilt gcc function` ` ` `// which will count total set bits in 'leftmask'` ` ` `if` `(__builtin_popcount(leftmask) == 1) {` ` ` `for` `(` `int` `i = 0; i < n; ++i) {` ` ` ` ` `// Since one person is present at left` ` ` `// side, thus return that person only` ` ` `if` `(leftmask & (1 << i)) {` ` ` `res = arr[i];` ` ` `break` `;` ` ` `}` ` ` `}` ` ` `}` ` ` `else` `{` ` ` ` ` `// try for every pair of people by` ` ` `// sending them to right side` ` ` ` ` `// Initialize the result with maximum value` ` ` `res = INT_MAX;` ` ` `for` `(` `int` `i = 0; i < n; ++i) {` ` ` ` ` `// If ith person is not present then` ` ` `// skip the rest loop` ` ` `if` `(!(leftmask & (1 << i)))` ` ` `continue` `;` ` ` ` ` `for` `(` `int` `j = i + 1; j < n; ++j) {` ` ` `if` `(leftmask & (1 << j)) {` ` ` ` ` `// Find maximum integer(slowest` ` ` `// person's time)` ` ` `int` `val = max(arr[i], arr[j]);` ` ` ` ` `// Recurse for other people after un-setting` ` ` `// the ith and jth bit of left-mask` ` ` `val += findMinTime(leftmask ^ (1 << i) ^ (1 << j),` ` ` `turn ^ 1, arr, n);` ` ` `// Find minimum answer among` ` ` `// all chosen values` ` ` `res = min(res, val);` ` ` `}` ` ` `}` ` ` `}` ` ` `}` ` ` `}` ` ` `return` `res;` `}` ` ` `// Utility function to find minimum time` `int` `findTime(` `int` `arr[], ` `int` `n)` `{` ` ` `// Find the mask of 'n' peoples` ` ` `int` `mask = (1 << n) - 1;` ` ` ` ` `// Initialize all entries in dp as -1` ` ` `memset` `(dp, -1, ` `sizeof` `(dp));` ` ` ` ` `return` `findMinTime(mask, 0, arr, n);` `}` ` ` `// Driver program` `int` `main()` `{` ` ` `int` `arr[] = { 10, 20, 30 };` ` ` `int` `n = ` `sizeof` `(arr)/` `sizeof` `(arr[0]);` ` ` `cout << findTime(arr, n);` ` ` `return` `0;` `}` |

Output60

**Time complexity: ****Auxiliary space: **

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