Minimum Total Distance Traveled | Leetcode 2463 | DP | RECURSION | Hard | Java | Developer Coder

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In this video, we will dive into the LeetCode problem Minimum Total Distance Traveled | Leetcode 2463. This problem challenges us with dynamic programming and recursion techniques to find the optimal solution. Join me as we explore the problem step-by-step using Java.

✨ What You Will Learn:

Dynamic Programming strategies
Recursive approaches
Efficient coding techniques in Java
Tips for tackling hard LeetCode problems
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[DeveloperCoder]

Recursion+Memo


class Solution {
public long robot, int[][] factory) {
Collections.sort(robot);
Arrays.sort(factory, Comparator.comparingInt(a -> a[0]));
List<Integer> factoryPositions = new ArrayList<>();
for(int[] f : factory ){
for(int i=0;i<f[1];i++){
factoryPositions.add(f[0]);
}
}
int robotCount = robot.size();
int factoryCount = factoryPositions.size();
long[][] memo = new
for(long[] row: memo){
Arrays.fill(row, -1);
}
return calculateMinDistance(0, 0, robot, factoryPositions, memo);
}
private long calculateMinDistance(int robotIdx, int factoryIdx, List<Integer> robot, List<Integer> factoryPositions, long[][] memo ){
//All robots assigned
if(robotIdx == robot.size()) return 0;
//No factories left to assign
if(factoryIdx == factoryPositions.size()) return (long) 1e12;
//check memo
if(memo[robotIdx][factoryIdx] != -1){
return memo[robotIdx][factoryIdx];
}

//option:1 assign current robot to current factory
long assign = Math.abs(robot.get(robotIdx) - + calculateMinDistance(robotIdx+1, factoryIdx+1, robot, factoryPositions, memo);

//option:2 Skip current factory for the current robot
long skip = calculateMinDistance(robotIdx, factoryIdx + 1, robot, factoryPositions, memo);

memo[robotIdx][factoryIdx] = Math.min(assign, skip);
return memo[robotIdx][factoryIdx];
}
}

[DeveloperCoder]

DP:

class Solution {
public long robot, int[][] factory) {
Collections.sort(robot);
Arrays.sort(factory, Comparator.comparingInt(a -> a[0]));
List<Integer> factoryPositions = new ArrayList<>();
for (int[] f : factory) {
for (int i = 0; i < f[1]; i++) {
factoryPositions.add(f[0]);
}
}
int robotCount = robot.size();
int factoryCount = factoryPositions.size();

long[] next = new long[factoryCount+1];
long[] current = new long[factoryCount+1];

for( int i=robotCount -1;i>=0;i--){
//no factories left case
if(i!= robotCount -1 ) next[factoryCount] = (long) 1e12;
current[factoryCount] = (long) 1e12;
for(int j = factoryCount -1 ;j>=0;j--){
//assign current robot to current factory
long assign = Math.abs((long) robot.get(i) - factoryPositions.get(j)) + next[j+1];
//skip current factory for this robot
long skip = current[j+1];
//take the min option
current[j] = Math.min(assign, skip);
}
System.arraycopy(current, 0, next, 0, factoryCount+1);
}
return current[0];

}
}

[arihantsinghrana]

can you provide the code
for both approaches