Palindrome Number - Practice Coding Problems

Palindrome Number - Problem

Math Easy

Given an integer x, return true if x is a palindrome, and false otherwise.

A palindrome number reads the same forward and backward. For example, 121 is a palindrome while 123 is not.

Follow up: Could you solve it without converting the integer to a string?

Input & Output

Example 1 — Positive Palindrome

$ Input: x = 121

› Output: true

💡 Note: 121 reads the same from left to right and right to left

Example 2 — Not a Palindrome

$ Input: x = -121

› Output: false

💡 Note: From left to right it reads -121, but from right to left it becomes 121- which is different

Example 3 — Single Digit

$ Input: x = 7

› Output: true

💡 Note: Single digit numbers are always palindromes

Constraints

-2³¹ ≤ x ≤ 2³¹ - 1

Visualization

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Understanding the Visualization

Input

Given integer x = 121

Process

Check if digits are symmetric

Output

Return true if palindrome, false otherwise

Key Takeaway

🎯 Key Insight: We only need to reverse half the number to check if it matches the other half

Asked in

a Amazon 15 M Microsoft 12 G Google 8 🍎 Apple 6

The key insight is that we can check palindromes mathematically without string conversion. The optimal approach reverses only half the digits to avoid overflow. Best approach: Half Reversal - Time: O(log n), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ String Conversion	O(log n)	O(log n)	Convert number to string and check if it reads the same forwards and backwards
Mathematical Reversal	O(log n)	O(1)	Reverse the entire number mathematically and compare with original
Half Reversal (Optimal)	O(log n)	O(1)	Reverse only half the digits to avoid overflow and improve efficiency

String Conversion — Algorithm Steps

Convert integer to string
Compare first and last characters, then move inward
Return false if any pair doesn't match

Visualization

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Step-by-Step Walkthrough

Convert to String

Transform number into string format

Two Pointers

Compare characters from start and end

Result

Return true if all pairs match

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <stdbool.h>

bool solution(int x) {
    if (x < 0) return false;
    
    char s[20];
    sprintf(s, "%d", x);
    int len = strlen(s);
    int left = 0, right = len - 1;
    
    while (left < right) {
        if (s[left] != s[right]) {
            return false;
        }
        left++;
        right--;
    }
    
    return true;
}

int main() {
    int x;
    scanf("%d", &x);
    
    bool result = solution(x);
    printf(result ? "true\n" : "false\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(log n)

Converting to string takes O(log n) time where log n is the number of digits

⚡ Linearithmic

Space Complexity

O(log n)

String conversion requires O(log n) space to store the digits

⚡ Linearithmic Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

String Conversion — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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