Palindrome Partitioning Ii

                                            
class Solution {
public:
    int minCut(string s) {
        int n = s.size();
        vector<vector<bool>> isPalindrome(n, vector<bool>(n, false));
        vector<int> dp(n, INT_MAX);

        for (int end = 0; end < n; end++) {
            for (int start = 0; start <= end; start++) {
                if (s[start] == s[end] && (end - start <= 2 || isPalindrome[start + 1][end - 1])) {
                    isPalindrome[start][end] = true;
                }
            }
        }

        for (int i = 0; i < n; i++) {
            if (isPalindrome[0][i]) {
                dp[i] = 0;
            } else {
                for (int j = 0; j < i; j++) {
                    if (isPalindrome[j + 1][i]) {
                        dp[i] = min(dp[i], dp[j] + 1);
                    }
                }
            }
        }

        return dp[n - 1];
    }
};

                                            
class Solution {
    public int minCut(String s) {
        int n = s.length();
        boolean[][] isPalindrome = new boolean[n][n];
        int[] dp = new int[n];
        
        for (int end = 0; end < n; end++) {
            dp[end] = end;
            for (int start = 0; start <= end; start++) {
                if (s.charAt(start) == s.charAt(end) && (end - start <= 2 || isPalindrome[start + 1][end - 1])) {
                    isPalindrome[start][end] = true;
                    dp[end] = (start == 0) ? 0 : Math.min(dp[end], dp[start - 1] + 1);
                }
            }
        }
        
        return dp[n - 1];
    }
}

                                            
class Solution(object):
    def minCut(self, s):
        n = len(s)
        is_palindrome = [[False] * n for _ in range(n)]
        
        for i in range(n):
            is_palindrome[i][i] = True
        
        for length in range(2, n + 1):
            for start in range(n - length + 1):
                end = start + length - 1
                if length == 2:
                    is_palindrome[start][end] = s[start] == s[end]
                else:
                    is_palindrome[start][end] = s[start] == s[end] and is_palindrome[start + 1][end - 1]
        
        dp = [float('inf')] * n
        for end in range(n):
            if is_palindrome[0][end]:
                dp[end] = 0
            else:
                dp[end] = min(dp[end], min(dp[i - 1] + 1 for i in range(end, 0, -1) if is_palindrome[i][end]))
        
        return dp[n - 1]

                                            
public class Solution {
    public int MinCut(string s) {
        int n = s.Length;
        bool[,] isPalindrome = new bool[n, n];
        int[] dp = new int[n];
        
        for (int i = 0; i < n; i++) {
            dp[i] = i;
            for (int j = 0; j <= i; j++) {
                if (s[j] == s[i] && (i - j < 2 || isPalindrome[j + 1, i - 1])) {
                    isPalindrome[j, i] = true;
                    dp[i] = (j == 0) ? 0 : Math.Min(dp[i], dp[j - 1] + 1);
                }
            }
        }
        
        return dp[n - 1];
    }
}

                                            
/**
 * @param {string} s
 * @return {number}
 */
var minCut = function(s) {
    const n = s.length;
    const isPalindrome = Array(n).fill(0).map(() => Array(n).fill(false));

    for (let i = n - 1; i >= 0; i--) {
        for (let j = i; j < n; j++) {
            if (s[i] === s[j] && (j - i <= 1 || isPalindrome[i + 1][j - 1])) {
                isPalindrome[i][j] = true;
            }
        }
    }

    const dp = Array(n).fill(Infinity);

    for (let i = 0; i < n; i++) {
        if (isPalindrome[0][i]) {
            dp[i] = 0;
        } else {
            for (let j = 0; j < i; j++) {
                if (isPalindrome[j + 1][i]) {
                    dp[i] = Math.min(dp[i], dp[j] + 1);
                }
            }
        }
    }

    return dp[n - 1];
};

                                            
function minCut(s: string): number {
    const n = s.length;
    const isPalindrome = Array.from({ length: n }, () => Array(n).fill(false));
    const dp = Array(n).fill(0);
    
    for (let i = 0; i < n; i++) {
        let minCuts = i;
        for (let j = 0; j <= i; j++) {
            if (s[i] === s[j] && (i - j <= 1 || isPalindrome[j + 1][i - 1])) {
                isPalindrome[j][i] = true;
                minCuts = j === 0 ? 0 : Math.min(minCuts, dp[j - 1] + 1);
            }
        }
        dp[i] = minCuts;
    }
    
    return dp[n - 1];
};

                                            
class Solution {
    /**
     * @param String $s
     * @return Integer
     */
    function minCut($s) {
        $n = strlen($s);
        $dp = array_fill(0, $n, 0);
        
        for ($i = 0; $i < $n; $i++) {
            $dp[$i] = $i;
        }
        
        for ($i = 0; $i < $n; $i++) {
            $this->checkPalindrome($s, $dp, $i, $i);
            $this->checkPalindrome($s, $dp, $i, $i + 1);
        }
        
        return $dp[$n - 1];
    }
    
    function checkPalindrome($s, &$dp, $left, $right) {
        $n = strlen($s);
        while ($left >= 0 && $right < $n && $s[$left] == $s[$right]) {
            $cut = ($left == 0) ? 0 : $dp[$left - 1] + 1;
            $dp[$right] = min($dp[$right], $cut);
            $left--;
            $right++;
        }
    }
}

                                            
class Solution {
    func minCut(_ s: String) -> Int {
        if s.isEmpty || s.count == 1 {
            return 0
        }
        
        let sArray = Array(s)
        let n = s.count
        var dp = Array(repeating: Array(repeating: false, count: n), count: n)
        var cuts = Array(repeating: 0, count: n)
        
        for j in 0..<n {
            var minCuts = j
            for i in 0...j {
                if sArray[i] == sArray[j] && (j - i <= 1 || dp[i+1][j-1]) {
                    dp[i][j] = true
                    minCuts = i == 0 ? 0 : min(minCuts, 1 + cuts[i - 1])
                }
            }
            cuts[j] = minCuts
        }
        
        return cuts[n-1]
    }
}

                                            
class Solution {
    fun minCut(s: String): Int {
        val n = s.length
        val dp = IntArray(n) { it }
        val isPalindrome = Array(n) { BooleanArray(n) }

        for (i in 0 until n) {
            for (j in 0..i) {
                if (s[i] == s[j] && (i - j <= 1 || isPalindrome[j + 1][i - 1])) {
                    isPalindrome[j][i] = true
                    if (j == 0) {
                        dp[i] = 0
                    } else {
                        dp[i] = minOf(dp[i], dp[j - 1] + 1)
                    }
                }
            }
        }

        return dp[n - 1]
    }
}

                                            
func minCut(s string) int {
    n := len(s)
    dp := make([][]bool, n)
    for i := range dp {
        dp[i] = make([]bool, n)
    }
    
    for length := 1; length <= n; length++ {
        for i := 0; i+length <= n; i++ {
            j := i + length - 1
            if length == 1 {
                dp[i][j] = true
            } else if length == 2 {
                dp[i][j] = s[i] == s[j]
            } else {
                dp[i][j] = s[i] == s[j] && dp[i+1][j-1]
            }
        }
    }
    
    cuts := make([]int, n)
    for i := range cuts {
        cuts[i] = i
    }
    
    for i := 0; i < n; i++ {
        if dp[0][i] {
            cuts[i] = 0
        } else {
            for j := 0; j < i; j++ {
                if dp[j+1][i] {
                    cuts[i] = min(cuts[i], cuts[j]+1)
                }
            }
        }
    }
    
    return cuts[n-1]
}

func min(a, b int) int {
    if a < b {
        return a
    }
    return b
}

                                            
# @param {String} s
# @return {Integer}
def min_cut(s)
    n = s.length
    is_palindrome = Array.new(n) { Array.new(n, false) }
    
    (0...n).each do |i|
        is_palindrome[i][i] = true
    end
    
    (2..n).each do |len|
        (0...n-len+1).each do |start|
            last = start + len - 1
            if len == 2
                is_palindrome[start][last] = s[start] == s[last]
            else
                is_palindrome[start][last] = s[start] == s[last] && is_palindrome[start+1][last-1]
            end
        end
    end
    
    cuts = Array.new(n)
    
    (0...n).each do |i|
        min_cut = i
        (0..i).each do |j|
            if is_palindrome[j][i]
                min_cut = 0 if j == 0
                min_cut = [min_cut, (j == 0 ? 0 : cuts[j-1]) + 1].min
            end
        end
        cuts[i] = min_cut
    end
    
    return cuts[n-1]
end

                                            
object Solution {
    def minCut(s: String): Int = {
        val n = s.length
        val isPalindrome = Array.ofDim[Boolean](n, n)
        val dp = Array.fill(n)(0)

        for (j <- 0 until n) {
            dp(j) = j
            for (i <- 0 to j) {
                if (s(i) == s(j) && (j - i <= 1 || isPalindrome(i + 1)(j - 1))) {
                    isPalindrome(i)(j) = true
                    dp(j) = if (i == 0) 0 else Math.min(dp(j), dp(i - 1) + 1)
                }
            }
        }

        dp(n - 1)
    }
}

                                            
impl Solution {
    pub fn min_cut(s: String) -> i32 {
        let n = s.len();
        let s = s.chars().collect::<Vec<_>>();
        let mut dp = vec![0; n];
        let mut pal = vec![vec![false; n]; n];
        
        for i in 0..n {
            let mut min = i as i32;
            for j in 0..=i {
                if s[i] == s[j] && (i - j <= 1 || pal[j + 1][i - 1]) {
                    pal[j][i] = true;
                    min = if j == 0 { 0 } else { min.min(dp[j - 1] + 1) };
                }
            }
            dp[i] = min;
        }
        
        dp[n - 1]
    }
}

To solve this coding challenge of finding the minimum cuts needed for palindrome partitioning of a given string

s

, we need to follow a systematic approach that breaks the problem into smaller, manageable subproblems. The key here is to use dynamic programming to efficiently keep track of palindromic substrings and calculate the minimum cuts required.

Explanation

Introduction

The challenge is to partition the string such that every substring of the partition is a palindrome and return the minimum number of cuts needed. Palindromes are sequences that read the same backward as forward, and the goal is to minimize the number of partitions that achieve this.

Dynamic Programming Approach

1. Precompute Palindromic Substrings

We first precompute whether a substring

s[i:j+1]

is a palindrome for all possible

i

and

j

. This involves maintaining a 2D list

is_palindrome

where

is_palindrome[i][j]

True

if the substring

s[i:j+1]

is a palindrome and

False

otherwise.

2. Initialize the DP Table

We then use another list

dp

where

dp[i]

represents the minimum number of cuts needed for the substring

s[0:i+1]

3. Populate the DP Table

The computation for

dp

involves iterating over the end indices of the substrings and calculating the minimum number of cuts needed based on previously computed values.

Detailed Steps in Pseudocode

Step 1: Initialize and Compute Palindrome Status

Create a 2D array
is_palindrome

of size
n x n

initialized to
False

.
All substrings of length 1 are palindromes.
Check substrings of length 2 and set palindromic status based on character equality.
Check longer substrings and use previously computed palindromic information for intermediate substrings.

                                            
# Define and initialize variables
n = length of s
is_palindrome = 2D array of size n x n initialized to False

# Every single character is a palindrome
for each i from 0 to n-1:
is_palindrome[i][i] = True

# Check for palindromes of length 2
for each i from 0 to n-2:
if s[i] == s[i+1]:
is_palindrome[i][i+1] = True

# Check for palindromes of length greater than 2
for length from 3 to n:
for start from 0 to n-length:
end = start + length - 1
if s[start] == s[end] and is_palindrome[start + 1][end - 1]:
is_palindrome[start][end] = True

Step 2: Compute Minimum Cuts Using DP

Initialize the
dp

array where
dp[i]

tracks the minimum cuts needed for the substring
s[0:i+1]

.
For each ending index, determine if the substring is already a palindrome; if yes, no cuts are required.
Otherwise, calculate the minimum number of cuts based on previous results and update the
dp

table accordingly.

                                            
# Initialize dp array
dp = array of size n with values initially set to infinity
for end in range 0 to n-1:
    if is_palindrome[0][end]:
        dp[end] = 0  # No cuts needed
    else:
        # Compute minimum cuts
        dp[end] = minimum of (dp[i - 1] + 1) for all i from 1 to end if is_palindrome[i][end]

Step 3: Return the Result

Return the value of

dp[n-1]

which gives the minimum cuts needed for partitioning the entire string

s

into palindromic substrings.

                                            
# Final answer
return dp[n-1]

Conclusion

By using dynamic programming, we efficiently compute the minimum cuts required for palindrome partitioning. The precomputation of palindromic substrings combined with a systematic update of the

dp

table ensures that we find the optimal solution in an efficient manner.