检查 substring[L…R] 是否为回文的查询

给定一个字符串str和Q查询。每个查询由两个数字L和R 组成。任务是打印子字符串 [L…R]是否是回文。

例子：

Input: str = “abacccde”, Q[][] = {{0, 2}, {1, 2}, {2, 4}, {3, 5}} 
Output: 
Yes 
No 
No 
Yes

Input: str = “abaaab”, Q[][] = {{0, 1}, {1, 5}} 
Output: 
No 
Yes  

简单的方法：一种简单的方法是检查每个子字符串是否为回文。在最坏的情况下，每个查询最多需要O(Q) 。

高效的方法： 一种有效的方法是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。
现在，对于每个查询，检查dp[l][r]是否为真。

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 answer every query in O(1)
void answerQuery(int l, int r, bool dp[N][N])
{
    if (dp[l][r])
        cout << "Yes\n";
    else
        cout << "No\n";
}
 
// Driver code
int main()
{
    string s = "abaaab";
    bool dp[N][N];
    pre_process(dp, s);
 
    int queries[][2] = { { 0, 1 }, { 1, 5 } };
    int q = sizeof(queries) / sizeof(queries[0]);
 
    for (int i = 0; i < q; i++)
        answerQuery(queries[i][0], queries[i][1], dp);
 
    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 answer every query in O(1)
    static void answerQuery(int l, int r, boolean dp[][])
    {
        if (dp[l][r]) {
            System.out.println("Yes");
        }
        else {
            System.out.println("No");
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String s = "abaaab";
        boolean[][] dp = new boolean[N][N];
        pre_process(dp, s.toCharArray());
 
        int queries[][] = { { 0, 1 }, { 1, 5 } };
        int q = queries.length;
 
        for (int i = 0; i < q; i++) {
            answerQuery(queries[i][0], queries[i][1], dp);
        }
    }
}
 
// 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 answer every query in O(1)
def answerQuery(l, r, dp):
 
    if (dp[l][r]):
        print("Yes")
    else:
        print("No")
 
# Driver code
s = "abaaab"
dp = [[0 for i in range(N)]
         for i in range(N)]
pre_process(dp, s)
 
queries = [[0, 1], [1, 5]]
q = len(queries)
 
for i in range(q):
    answerQuery(queries[i][0],
                queries[i][1], dp)
 
# This code is contributed by mohit kumar

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 answer every query in O(1)
    static void answerQuery(int l, int r, bool[, ] dp)
    {
        if (dp[l, r]) {
            Console.WriteLine("Yes");
        }
        else {
            Console.WriteLine("No");
        }
    }
 
    // Driver code
    public static void Main()
    {
        string s = "abaaab";
        bool[, ] dp = new bool[N, N];
        pre_process(dp, s.ToCharArray());
 
        int[, ] queries = { { 0, 1 }, { 1, 5 } };
        int q = queries.Length;
 
        for (int i = 0; i < q; i++) {
            answerQuery(queries[i, 0], queries[i, 1], dp);
        }
    }
}
 
// This code is contributed by Ankit.

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 answer every query in O(1)
    function answerQuery(l , r,  dp) {
        if (dp[l][r]) {
            document.write("Yes<br/>");
        }
        else {
            document.write("No<br/>");
        }
    }
 
    // Driver code
     
    var s = "abaaab";
    var dp = Array(N).fill().map(()=>Array(N).fill());
    pre_process(dp, s);
 
    var queries = [ [ 0, 1 ], [ 1, 5 ] ];
    var q = queries.length;
 
    for (i = 0; i < q; i++) {
        answerQuery(queries[i][0], queries[i][1], dp);
    }
 
// This code contributed by Rajput-Ji

输出：

No
Yes

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