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: 60 Explanation 1. 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 = 60 Input: Crossing Times = [1, 2, 5, 8} Output: 15 Explanation See 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; } |
Output 60
Time complexity:
Auxiliary space:
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