LeetCode [1320] Minimum Distance to Type a Word Using Two Fingers

 1320. Minimum Distance to Type a Word Using Two Fingers

Hard

You have a keyboard layout as shown above in the XY plane, where each English uppercase letter is located at some coordinate, for example, the letter A is located at coordinate (0,0), the letter B is located at coordinate (0,1), the letter P is located at coordinate (2,3) and the letter Z is located at coordinate (4,1).

Given the string word, return the minimum total distance to type such string using only two fingers. The distance between coordinates (x1,y1) and (x2,y2) is |x1 - x2| + |y1 - y2|. 

Note that the initial positions of your two fingers are considered free so don't count towards your total distance, also your two fingers do not have to start at the first letter or the first two letters.

 

Example 1:

Input: word = "CAKE"
Output: 3
Explanation: 
Using two fingers, one optimal way to type "CAKE" is: 
Finger 1 on letter 'C' -> cost = 0 
Finger 1 on letter 'A' -> cost = Distance from letter 'C' to letter 'A' = 2 
Finger 2 on letter 'K' -> cost = 0 
Finger 2 on letter 'E' -> cost = Distance from letter 'K' to letter 'E' = 1 
Total distance = 3

Example 2:

Input: word = "HAPPY"
Output: 6
Explanation: 
Using two fingers, one optimal way to type "HAPPY" is:
Finger 1 on letter 'H' -> cost = 0
Finger 1 on letter 'A' -> cost = Distance from letter 'H' to letter 'A' = 2
Finger 2 on letter 'P' -> cost = 0
Finger 2 on letter 'P' -> cost = Distance from letter 'P' to letter 'P' = 0
Finger 1 on letter 'Y' -> cost = Distance from letter 'A' to letter 'Y' = 4
Total distance = 6

Example 3:

Input: word = "NEW"
Output: 3

Example 4:

Input: word = "YEAR"
Output: 7

 

Constraints:

• 2 <= word.length <= 300
• Each word[i] is an English uppercase letter.

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class Solution { String word; int[][][] dp; int n; int cost(int a, int b){ if(a==0) return 0; a--; b--; int ax = a/6, ay = a%6; int bx = b/6, by = b%6; return Math.abs(ax-bx) + Math.abs(ay-by); } public int minimumDistance(String word) { this.word = word; this.n = word.length(); //dp[i][j][p] is when left finger is at i and right finger is at j, the distance takes to complete word[p:] dp = new int[27][27][300]; return dfs(0, 0, 0); } //compute min distance to type word[wPos:] starting from [lPos, rPos] int dfs(int wPos, int lPos, int rPos){ if(wPos>=n) return 0; if(dp[lPos][rPos][wPos]>0) return dp[lPos][rPos][wPos]; int next = word.charAt(wPos)-'A'+1; //if type next by left finger int l = cost(lPos, next) + dfs(wPos+1, next, rPos); //if type next by right finger int r = cost(rPos, next) + dfs(wPos+1, lPos, next); dp[lPos][rPos][wPos] = Math.min(l, r); return dp[lPos][rPos][wPos]; } }