String Reversal और Palindrome

किसी string को reverse करना एक word को उल्टा पढ़ने जैसा है, और एक palindrome level जैसा एक word है जो दोनों ends से एक जैसा दिखता है।

In this page:

String Reversal
Palindrome Basics
Two Pointer Method
Ignoring Case
Palindrome Practice

Syntax

markup

reversed_text = text[::-1]

left, right = 0, len(text) - 1
while left < right:
    if text[left] != text[right]:
        return False
    left += 1
    right -= 1
return True

String Reversal

किसी string को reverse करने का मतलब है उन्हीं characters के साथ opposite order में एक नई string produce करना, आखिरी character से पहले तक, जो आखिर से पीछे की ओर एक नई string बनाकर या characters को जगह पर swap करके किया जा सकता है।

उदाहरण: String Reversal

#include <iostream>
#include <algorithm>
using namespace std;
int main() {
    string s = "hello";
    reverse(s.begin(), s.end());
    cout << "Reversed: " << s << endl;
    return 0;
}

public class Main {
    public static void main(String[] args) {
        String s = "hello";
        String reversed = new StringBuilder(s).reverse().toString();
        System.out.println("Reversed: " + reversed);
    }
}

s = "hello"
print("Reversed:", s[::-1])

#include <stdio.h>
#include <string.h>
int main() {
    char s[] = "hello";
    int n = strlen(s);
    for (int i = 0; i < n / 2; i++) {
        char t = s[i]; s[i] = s[n - 1 - i]; s[n - 1 - i] = t;
    }
    printf("Reversed: %s\n", s);
    return 0;
}

Palindrome Basics

एक palindrome एक string है जो आगे और पीछे दोनों तरह से identically पढ़ती है, जैसे level या racecar। एक की जांच करने का मतलब है string की तुलना इसके खुद के reverse से करना, या दोनों ends से अंदर की ओर characters compare करना।

उदाहरण: Palindrome Basics

#include <iostream>
#include <algorithm>
using namespace std;
int main() {
    string s = "racecar";
    string rev = s;
    reverse(rev.begin(), rev.end());
    cout << (s == rev ? "Palindrome" : "Not a palindrome") << endl;
    return 0;
}

public class Main {
    public static void main(String[] args) {
        String s = "racecar";
        String rev = new StringBuilder(s).reverse().toString();
        System.out.println(s.equals(rev) ? "Palindrome" : "Not a palindrome");
    }
}

s = "racecar"
print("Palindrome" if s == s[::-1] else "Not a palindrome")

#include <stdio.h>
#include <string.h>
int main() {
    char s[] = "racecar";
    int n = strlen(s), isPal = 1;
    for (int i = 0; i < n / 2; i++) if (s[i] != s[n - 1 - i]) isPal = 0;
    printf("%s\n", isPal ? "Palindrome" : "Not a palindrome");
    return 0;
}

Two Pointer Method

Two-pointer method कभी एक reversed copy बनाए बिना एक palindrome जांचती है: एक pointer front से शुरू होता है, एक back से, और वे middle की ओर move करते हुए characters compare करते हैं, एक mismatch मिलते ही जल्दी रुकते हुए।

उदाहरण: Two Pointer Method

#include <iostream>
using namespace std;
int main() {
    string s = "level";
    int left = 0, right = s.length() - 1;
    bool isPal = true;
    while (left < right) {
        if (s[left] != s[right]) { isPal = false; break; }
        left++; right--;
    }
    cout << (isPal ? "Palindrome" : "Not a palindrome") << endl;
    return 0;
}

public class Main {
    public static void main(String[] args) {
        String s = "level";
        int left = 0, right = s.length() - 1;
        boolean isPal = true;
        while (left < right) {
            if (s.charAt(left) != s.charAt(right)) { isPal = false; break; }
            left++; right--;
        }
        System.out.println(isPal ? "Palindrome" : "Not a palindrome");
    }
}

s = "level"
left, right = 0, len(s) - 1
is_pal = True
while left < right:
    if s[left] != s[right]:
        is_pal = False
        break
    left += 1; right -= 1
print("Palindrome" if is_pal else "Not a palindrome")

#include <stdio.h>
#include <string.h>
int main() {
    char s[] = "level";
    int left = 0, right = strlen(s) - 1, isPal = 1;
    while (left < right) {
        if (s[left] != s[right]) { isPal = 0; break; }
        left++; right--;
    }
    printf("%s\n", isPal ? "Palindrome" : "Not a palindrome");
    return 0;
}

Ignoring Case

एक palindrome check को case-insensitive बनाने के लिए, compare हो रहे दोनों characters को compare करने से पहले एक जैसे case (आमतौर पर lowercase) में convert करें, ताकि Level को अब भी सही से एक palindrome पहचाना जाए।

उदाहरण: Ignoring Case

#include <iostream>
#include <cctype>
using namespace std;
int main() {
    string s = "Level";
    int left = 0, right = s.length() - 1;
    bool isPal = true;
    while (left < right) {
        if (tolower(s[left]) != tolower(s[right])) { isPal = false; break; }
        left++; right--;
    }
    cout << (isPal ? "Palindrome (case-insensitive)" : "Not a palindrome") << endl;
    return 0;
}

public class Main {
    public static void main(String[] args) {
        String s = "Level";
        int left = 0, right = s.length() - 1;
        boolean isPal = true;
        while (left < right) {
            if (Character.toLowerCase(s.charAt(left)) != Character.toLowerCase(s.charAt(right))) { isPal = false; break; }
            left++; right--;
        }
        System.out.println(isPal ? "Palindrome (case-insensitive)" : "Not a palindrome");
    }
}

s = "Level"
left, right = 0, len(s) - 1
is_pal = True
while left < right:
    if s[left].lower() != s[right].lower():
        is_pal = False
        break
    left += 1; right -= 1
print("Palindrome (case-insensitive)" if is_pal else "Not a palindrome")

#include <stdio.h>
#include <string.h>
#include <ctype.h>
int main() {
    char s[] = "Level";
    int left = 0, right = strlen(s) - 1, isPal = 1;
    while (left < right) {
        if (tolower(s[left]) != tolower(s[right])) { isPal = 0; break; }
        left++; right--;
    }
    printf("%s\n", isPal ? "Palindrome (case-insensitive)" : "Not a palindrome");
    return 0;
}

Palindrome Practice

Palindrome problems careful index management practice करने का एक बढ़िया तरीका हैं, क्योंकि string की boundaries पर off-by-one errors सबसे आम bug हैं, खासकर non-alphanumeric characters भी skip करते समय।

उदाहरण: Palindrome Practice

#include <iostream>
#include <cctype>
using namespace std;
int main() {
    string s = "A man a plan a canal Panama";
    int left = 0, right = s.length() - 1;
    bool isPal = true;
    while (left < right) {
        if (!isalnum(s[left])) { left++; continue; }
        if (!isalnum(s[right])) { right--; continue; }
        if (tolower(s[left]) != tolower(s[right])) { isPal = false; break; }
        left++; right--;
    }
    cout << (isPal ? "Palindrome" : "Not a palindrome") << endl;
    return 0;
}

public class Main {
    public static void main(String[] args) {
        String s = "A man a plan a canal Panama";
        int left = 0, right = s.length() - 1;
        boolean isPal = true;
        while (left < right) {
            if (!Character.isLetterOrDigit(s.charAt(left))) { left++; continue; }
            if (!Character.isLetterOrDigit(s.charAt(right))) { right--; continue; }
            if (Character.toLowerCase(s.charAt(left)) != Character.toLowerCase(s.charAt(right))) { isPal = false; break; }
            left++; right--;
        }
        System.out.println(isPal ? "Palindrome" : "Not a palindrome");
    }
}

s = "A man a plan a canal Panama"
left, right = 0, len(s) - 1
is_pal = True
while left < right:
    if not s[left].isalnum():
        left += 1; continue
    if not s[right].isalnum():
        right -= 1; continue
    if s[left].lower() != s[right].lower():
        is_pal = False
        break
    left += 1; right -= 1
print("Palindrome" if is_pal else "Not a palindrome")

#include <stdio.h>
#include <string.h>
#include <ctype.h>
int main() {
    char s[] = "A man a plan a canal Panama";
    int left = 0, right = strlen(s) - 1, isPal = 1;
    while (left < right) {
        if (!isalnum(s[left])) { left++; continue; }
        if (!isalnum(s[right])) { right--; continue; }
        if (tolower(s[left]) != tolower(s[right])) { isPal = 0; break; }
        left++; right--;
    }
    printf("%s\n", isPal ? "Palindrome" : "Not a palindrome");
    return 0;
}

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