Day 30

Unique Paths

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class Solution {
    public int uniquePaths(int m, int n) {
        int[][] dp = new int[m][n];
        for (int i = 0; i < m; i++) {
            dp[i][0] = 1;
        }
        for (int i = 0; i < n; i++) {
            dp[0][i] = 1;
        }
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }
        }
        return dp[m - 1][n - 1];
    }
}
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func uniquePaths(m int, n int) int {
	dp := make([][]int, m)
	for i := range dp {
		dp[i] = make([]int, n)
		dp[i][0] = 1
	}
	for j := 0; j < n; j++ {
		dp[0][j] = 1
	}
	for i := 1; i < m; i++ {
		for j := 1; j < n; j++ {
			dp[i][j] = dp[i-1][j] + dp[i][j-1]
		}
	}
	return dp[m-1][n-1]
}

Unique Paths II

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class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        int[][] dp = new int[m][n];
        for (int i = 0; i < m; i++) {
            if (obstacleGrid[i][0] == 1) {
                break;
            }
            dp[i][0] = 1;
        }
        for (int i = 0; i < n; i++) {
            if (obstacleGrid[0][i] == 1) {
                break;
            }
            dp[0][i] = 1;
        }
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (obstacleGrid[i][j] == 1) {
                    dp[i][j] = 0;
                    continue;
                }
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }
        }
        return dp[m - 1][n - 1];
    }
}
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func uniquePathsWithObstacles(obstacleGrid [][]int) int {
	m, n := len(obstacleGrid), len(obstacleGrid[0])
	dp := make([][]int, m)
	for i := range dp {
		dp[i] = make([]int, n)
	}
	for i := 0; i < m && obstacleGrid[i][0] != 1; i++ {
		dp[i][0] = 1
	}
	for i := 0; i < n && obstacleGrid[0][i] != 1; i++ {
		dp[0][i] = 1
	}

	for i := 1; i < m; i++ {
		for j := 1; j < n; j++ {
			if obstacleGrid[i][j] == 1 {
				continue
			}
			dp[i][j] = dp[i-1][j] + dp[i][j-1]
		}
	}
	return dp[m-1][n-1]
}

Integer Break

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class Solution {
    public int integerBreak(int n) {
        int[] dp = new int[n+1];
        dp[2] = 1;
        for (int i = 3; i <= n; i++) {
            for (int j = 1; j <= i - j; j++) {
                dp[i] = Math.max(dp[i], Math.max(j* (i - j), j * dp[i-j]));
            }
        }
        return dp[n];
    }
}
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func integerBreak(n int) int {
	dp := make([]int, n+1)
	dp[2] = 1
	for i := 3; i <= n; i++ {
		for j := 1; j <= i-j; j++ {
			dp[i] = max(dp[i], max(j*(i-j), dp[i-j]*j))
		}
	}
	return dp[n]
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}