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leetcode_java/src/main/java/LeetCodeJava/DynamicProgramming/UniquePaths.java
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| package LeetCodeJava.DynamicProgramming; | ||
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| // https://leetcode.com/problems/unique-paths/description/ | ||
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| import java.math.BigInteger; | ||
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| public class UniquePaths { | ||
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| // VO | ||
| // IDEA : MATH | ||
| // -> the UNIQUE combination of x "a", and y "b" | ||
| // -> e.g. [a, a,....a] and [b,b...,,,,b] | ||
| // <- x count -> <- y count -> | ||
| // -> so the combination count is | ||
| // (x+y)! / (x! * y!) | ||
| public int uniquePaths(int m, int n) { | ||
| if (m == 0 || n == 0) { | ||
| return 0; | ||
| } | ||
| if (m == 1 || n == 1) { | ||
| return 1; | ||
| } | ||
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| /** NOTE !!! BigInteger op code */ | ||
| BigInteger res = getFactorial((m - 1) + (n - 1)) | ||
| .divide(getFactorial(m - 1).multiply(getFactorial(n - 1))); | ||
| return res.intValue(); | ||
| } | ||
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| private BigInteger getFactorial(int x) { | ||
| if (x <= 0) { | ||
| throw new ArithmeticException("x should be equal or bigger than 1"); | ||
| } | ||
| /** NOTE !!! BigInteger op code */ | ||
| BigInteger res = BigInteger.ONE; | ||
| for (int i = 1; i < x + 1; i++) { | ||
| res = res.multiply(BigInteger.valueOf(i)); | ||
| } | ||
| return res; | ||
| } | ||
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| // V1 | ||
| // https://leetcode.com/problems/unique-paths/solutions/4795217/memoization-and-tabulation-java-100-beats/ | ||
| public int uniquePaths_1(int m, int n) { | ||
| Integer[][] memo = new Integer[m][n]; | ||
| return findPath(m - 1, n - 1, memo); | ||
| } | ||
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| private int findPath(int r, int c, Integer[][] memo){ | ||
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| if(r == 0 && c == 0) | ||
| return 1; | ||
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| if(r < 0 || c < 0) | ||
| return 0; | ||
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| if(memo[r][c] != null) | ||
| return memo[r][c]; | ||
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| int up = findPath(r - 1, c, memo); | ||
| int left = findPath(r, c - 1, memo); | ||
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| memo[r][c] = up + left; | ||
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| return memo[r][c]; | ||
| } | ||
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| // V2 | ||
| // https://leetcode.com/problems/unique-paths/solutions/4801294/recurrsion-memoization-tabulation-easy-explaination/ | ||
| // Recurrsion | ||
| // This is a reccursive solution and will give TLE with reccursive solution, try these with DP | ||
| // public int uniquePaths(int m, int n) { | ||
| // return f(m, n, m-1, n-1); | ||
| // } | ||
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| // public int f(int m, int n, int r, int c){ | ||
| // if(r == || c == n-1){ | ||
| // return 1; | ||
| // } | ||
| // if(r < 0 || c < 0){ | ||
| // return 0; | ||
| // } | ||
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| // int up= f(m, n, r-1, c); | ||
| // int left= f(m, n, r, c-1); | ||
| // return up + left; | ||
| // } | ||
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| //Memoization | ||
| // public int uniquePaths(int m, int n) { | ||
| // int[][] dp= new int[m+1][n+1]; | ||
| // for (int[] row : dp) { | ||
| // Arrays.fill(row, -1); // Initialize each cell of the array individually | ||
| // } | ||
| // return f(m, n, m-1, n-1, dp); | ||
| // } | ||
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| // public int f(int m, int n, int r, int c, int[][] dp){ | ||
| // if(r == 0 || c == 0){ | ||
| // return 1; | ||
| // } | ||
| // if(r < 0 || c < 0){ | ||
| // return 0; | ||
| // } | ||
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| // if(dp[r][c] != -1){ | ||
| // return dp[r][c]; | ||
| // } | ||
| // int up= f(m, n, r-1, c, dp); | ||
| // int left= f(m, n, r, c-1, dp); | ||
| // return dp[r][c]= up + left; | ||
| // } | ||
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| // Tabulation | ||
| public int uniquePaths_2(int m, int n) { | ||
| int[][] dp= new int[m+1][n+1]; | ||
| dp[0][0]= 1; | ||
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| for(int i=0; i< m; i++){ | ||
| for(int j=0; j< n; j++){ | ||
| if(i==0 && j==0) continue; | ||
| int up= 0; | ||
| int left=0; | ||
| if(i>0) up= dp[i-1][j]; | ||
| if(j>0) left= dp[i][j-1]; | ||
| dp[i][j]= up + left; | ||
| } | ||
| } | ||
| return dp[m-1][n-1]; | ||
| } | ||
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| } |