计算给定字符串的非重叠回文子字符串对

给定一个字符串S 。任务是计算回文子串S1和S2的非重叠对，使得字符串应该是S1[L1…R1]和S2[L2…R2] ，其中0 ≤ L1 ≤ R1 < L2 ≤ R2 < N。任务是计算非重叠回文子串的对数。
例子：

Input: s = “aaa” 
Output: 5 
All possible pairs are (s[0], s[1]), (s[0], s[2]), 
(s[0], s[1, 2]), (s[1], s[2]) and (s[0, 1], s[2])
Input: s = “abacaba” 
Output: 36 
 

方法：我们可以使用动态规划来解决上述问题。我们可以最初创建DP表，该表存储substring[i….j]是否为回文。我们维护一个以自下而上的方式填充的布尔值dp[n][n] 。如果子串是回文， dp[i][j] 的值为真，否则为假。要计算 dp[i][j]，我们首先检查dp[i+1][j-1]的值，如果该值为真并且s[i]与s[j]相同，那么我们使dp [i][j]正确。否则， dp[i][j] 的值为假。此后可以按照以下步骤获取对数。

• 创建一个left[]数组，其中left[i]存储索引 i 左侧回文数的计数，包括 i。
• 创建一个right[]数组，其中right[i]存储索引 i 右侧的回文数，包括 i。
• 从0迭代到length-1并添加left[i]*right[i+1] 。每个索引的总和将是所需的对数。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
#define N 100
 
// Pre-processing function
void pre_process(bool dp[N][N], string s)
{
 
    // Get the size of the string
    int n = s.size();
 
    // Initially mark every
    // position as false
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++)
            dp[i][j] = false;
    }
 
    // For the length
    for (int j = 1; j <= n; j++) {
 
        // Iterate for every index with
        // length j
        for (int i = 0; i <= n - j; i++) {
 
            // If the length is less than 2
            if (j <= 2) {
 
                // If characters are equal
                if (s[i] == s[i + j - 1])
                    dp[i][i + j - 1] = true;
            }
 
            // Check for equal
            else if (s[i] == s[i + j - 1])
                dp[i][i + j - 1] = dp[i + 1][i + j - 2];
        }
    }
}
 
// Function to return the number of pairs
int countPairs(string s)
{
 
    // Create the dp table initially
    bool dp[N][N];
    pre_process(dp, s);
    int n = s.length();
 
    // Declare the left array
    int left[n];
    memset(left, 0, sizeof left);
 
    // Declare the right array
    int right[n];
    memset(right, 0, sizeof right);
 
    // Initially left[0] is 1
    left[0] = 1;
 
    // Count the number of palindrome
    // pairs to the left
    for (int i = 1; i < n; i++) {
 
        for (int j = 0; j <= i; j++) {
 
            if (dp[j][i] == 1)
                left[i]++;
        }
    }
 
    // Initially right most as 1
    right[n - 1] = 1;
 
    // Count the number of palindrome
    // pairs to the right
    for (int i = n - 2; i >= 0; i--) {
 
        right[i] = right[i + 1];
 
        for (int j = n - 1; j >= i; j--) {
 
            if (dp[i][j] == 1)
                right[i]++;
        }
    }
 
    int ans = 0;
 
    // Count the number of pairs
    for (int i = 0; i < n - 1; i++)
        ans += left[i] * right[i + 1];
 
    return ans;
}
 
// Driver code
int main()
{
    string s = "abacaba";
    cout << countPairs(s);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
 
    static int N = 100;
 
    // Pre-processing function
    static void pre_process(boolean dp[][], char[] s)
    {
 
        // Get the size of the string
        int n = s.length;
 
        // Initially mark every
        // position as false
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
            {
                dp[i][j] = false;
            }
        }
 
        // For the length
        for (int j = 1; j <= n; j++)
        {
 
            // Iterate for every index with
            // length j
            for (int i = 0; i <= n - j; i++)
            {
 
                // If the length is less than 2
                if (j <= 2)
                {
 
                    // If characters are equal
                    if (s[i] == s[i + j - 1])
                    {
                        dp[i][i + j - 1] = true;
                    }
                }
                // Check for equal
                else if (s[i] == s[i + j - 1])
                {
                    dp[i][i + j - 1] = dp[i + 1][i + j - 2];
                }
            }
        }
    }
 
    // Function to return the number of pairs
    static int countPairs(String s)
    {
 
        // Create the dp table initially
        boolean dp[][] = new boolean[N][N];
        pre_process(dp, s.toCharArray());
        int n = s.length();
 
        // Declare the left array
        int left[] = new int[n];
 
        // Declare the right array
        int right[] = new int[n];
 
        // Initially left[0] is 1
        left[0] = 1;
 
        // Count the number of palindrome
        // pairs to the left
        for (int i = 1; i < n; i++)
        {
 
            for (int j = 0; j <= i; j++)
            {
 
                if (dp[j][i] == true)
                {
                    left[i]++;
                }
            }
        }
 
        // Initially right most as 1
        right[n - 1] = 1;
 
        // Count the number of palindrome
        // pairs to the right
        for (int i = n - 2; i >= 0; i--)
        {
 
            right[i] = right[i + 1];
 
            for (int j = n - 1; j >= i; j--)
            {
 
                if (dp[i][j] == true)
                {
                    right[i]++;
                }
            }
        }
 
        int ans = 0;
 
        // Count the number of pairs
        for (int i = 0; i < n - 1; i++)
        {
            ans += left[i] * right[i + 1];
        }
 
        return ans;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String s = "abacaba";
        System.out.println(countPairs(s));
    }
}
 
// This code contributed by Rajput-Ji

Python3

# Python3 implementation of the approach
N = 100
 
# Pre-processing function
def pre_process(dp, s):
 
    # Get the size of the string
    n = len(s)
 
    # Initially mark every
    # position as false
    for i in range(n):
        for j in range(n):
            dp[i][j] = False
 
    # For the length
    for j in range(1, n + 1):
 
        # Iterate for every index with
        # length j
        for i in range(n - j + 1):
 
            # If the length is less than 2
            if (j <= 2):
 
                # If characters are equal
                if (s[i] == s[i + j - 1]):
                    dp[i][i + j - 1] = True
 
            # Check for equal
            elif (s[i] == s[i + j - 1]):
                dp[i][i + j - 1] = dp[i + 1][i + j - 2]
 
# Function to return the number of pairs
def countPairs(s):
 
    # Create the dp table initially
    dp = [[False for i in range(N)]
                 for j in range(N)]
    pre_process(dp, s)
    n = len(s)
 
    # Declare the left array
    left = [0 for i in range(n)]
 
    # Declare the right array
    right = [0 for i in range(n)]
 
    # Initially left[0] is 1
    left[0] = 1
 
    # Count the number of palindrome
    # pairs to the left
    for i in range(1, n):
 
        for j in range(i + 1):
 
            if (dp[j][i] == 1):
                left[i] += 1
 
    # Initially right most as 1
    right[n - 1] = 1
 
    # Count the number of palindrome
    # pairs to the right
    for i in range(n - 2, -1, -1):
 
        right[i] = right[i + 1]
 
        for j in range(n - 1, i - 1, -1):
 
            if (dp[i][j] == 1):
                right[i] += 1
 
    ans = 0
 
    # Count the number of pairs
    for i in range(n - 1):
        ans += left[i] * right[i + 1]
 
    return ans
 
# Driver code
s = "abacaba"
print(countPairs(s))
 
# This code is contributed by mohit kumar

PHP

= 0; $i--)
    {
        $right[$i] = $right[$i + 1];
 
        for ($j = $n - 1; $j >= $i; $j--)
        {
            if ($dp[$i][$j] == 1)
                $right[$i]++;
        }
    }
 
    $ans = 0;
 
    // Count the number of pairs
    for ($i = 0; $i < $n - 1; $i++)
        $ans += $left[$i] * $right[$i + 1];
 
    return $ans;
}
 
// Driver code
$s = "abacaba";
echo countPairs($s);
 
// This code is contributed by Ryuga
?>

C#

// C# implementation of the approach
using System;
     
class GFG
{
 
    static int N = 100;
 
    // Pre-processing function
    static void pre_process(bool [,]dp, char[] s)
    {
 
        // Get the size of the string
        int n = s.Length;
 
        // Initially mark every
        // position as false
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
            {
                dp[i,j] = false;
            }
        }
 
        // For the length
        for (int j = 1; j <= n; j++)
        {
 
            // Iterate for every index with
            // length j
            for (int i = 0; i <= n - j; i++)
            {
 
                // If the length is less than 2
                if (j <= 2)
                {
 
                    // If characters are equal
                    if (s[i] == s[i + j - 1])
                    {
                        dp[i,i + j - 1] = true;
                    }
                }
                // Check for equal
                else if (s[i] == s[i + j - 1])
                {
                    dp[i,i + j - 1] = dp[i + 1,i + j - 2];
                }
            }
        }
    }
 
    // Function to return the number of pairs
    static int countPairs(String s)
    {
 
        // Create the dp table initially
        bool [,]dp = new bool[N,N];
        pre_process(dp, s.ToCharArray());
        int n = s.Length;
 
        // Declare the left array
        int []left = new int[n];
 
        // Declare the right array
        int []right = new int[n];
 
        // Initially left[0] is 1
        left[0] = 1;
 
        // Count the number of palindrome
        // pairs to the left
        for (int i = 1; i < n; i++)
        {
 
            for (int j = 0; j <= i; j++)
            {
 
                if (dp[j,i] == true)
                {
                    left[i]++;
                }
            }
        }
 
        // Initially right most as 1
        right[n - 1] = 1;
 
        // Count the number of palindrome
        // pairs to the right
        for (int i = n - 2; i >= 0; i--)
        {
 
            right[i] = right[i + 1];
 
            for (int j = n - 1; j >= i; j--)
            {
 
                if (dp[i,j] == true)
                {
                    right[i]++;
                }
            }
        }
 
        int ans = 0;
 
        // Count the number of pairs
        for (int i = 0; i < n - 1; i++)
        {
            ans += left[i] * right[i + 1];
        }
 
        return ans;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        String s = "abacaba";
        Console.Write(countPairs(s));
    }
}
 
/* This code contributed by PrinciRaj1992 */

Javascript

// javascript implementation of the approach   
var N = 100;
 
    // Pre-processing function
    function pre_process( dp,  s) {
 
        // Get the size of the string
        var n = s.length;
 
        // Initially mark every
        // position as false
        for (i = 0; i < n; i++) {
            for (j = 0; j < n; j++) {
                dp[i][j] = false;
            }
        }
 
        // For the length
        for (j = 1; j <= n; j++) {
 
            // Iterate for every index with
            // length j
            for (i = 0; i <= n - j; i++) {
 
                // If the length is less than 2
                if (j <= 2) {
 
                    // If characters are equal
                    if (s[i] == s[i + j - 1]) {
                        dp[i][i + j - 1] = true;
                    }
                }
                // Check for equal
                else if (s[i] == s[i + j - 1]) {
                    dp[i][i + j - 1] = dp[i + 1][i + j - 2];
                }
            }
        }
    }
 
    // Function to return the number of pairs
    function countPairs(s) {
 
        // Create the dp table initially
        var dp = Array(N).fill().map(()=>Array(N).fill(false));
        pre_process(dp, s);
        var n = s.length;
 
        // Declare the left array
        var left = Array(n).fill(0);
 
        // Declare the right array
        var right = Array(n).fill(0);
 
        // Initially left[0] is 1
        left[0] = 1;
 
        // Count the number of palindrome
        // pairs to the left
        for (i = 1; i < n; i++) {
 
            for (j = 0; j <= i; j++) {
 
                if (dp[j][i] == true) {
                    left[i]++;
                }
            }
        }
 
        // Initially right most as 1
        right[n - 1] = 1;
 
        // Count the number of palindrome
        // pairs to the right
        for (i = n - 2; i >= 0; i--) {
 
            right[i] = right[i + 1];
 
            for (j = n - 1; j >= i; j--) {
 
                if (dp[i][j] == true) {
                    right[i]++;
                }
            }
        }
 
        var ans = 0;
 
        // Count the number of pairs
        for (i = 0; i < n - 1; i++) {
            ans += left[i] * right[i + 1];
        }
 
        return ans;
    }
 
    // Driver code
     
        var s = "abacaba";
        document.write(countPairs(s));
 
// This code is contributed by todaysgaurav

输出：

36

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