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2443. Sum of Number and Its Reverse

Description

Given a non-negative integer num, return true if num can be expressed as the sum of any non-negative integer and its reverse, or false otherwise.

 

Example 1:

Input: num = 443
Output: true
Explanation: 172 + 271 = 443 so we return true.

Example 2:

Input: num = 63
Output: false
Explanation: 63 cannot be expressed as the sum of a non-negative integer and its reverse so we return false.

Example 3:

Input: num = 181
Output: true
Explanation: 140 + 041 = 181 so we return true. Note that when a number is reversed, there may be leading zeros.

 

Constraints:

• 0 <= num <= 105

Solutions

Solution 1: Brute Force Enumeration

Enumerate $k$ in the range $[0,.., num]$, and check whether $k + reverse(k)$ equals $num$.

The time complexity is $O(n \times \log n)$, where $n$ is the size of $num$.

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• class Solution {
      public boolean sumOfNumberAndReverse(int num) {
          for (int x = 0; x <= num; ++x) {
              int k = x;
              int y = 0;
              while (k > 0) {
                  y = y * 10 + k % 10;
                  k /= 10;
              }
              if (x + y == num) {
                  return true;
              }
          }
          return false;
      }
  }
  

• class Solution {
  public:
      bool sumOfNumberAndReverse(int num) {
          for (int x = 0; x <= num; ++x) {
              int k = x;
              int y = 0;
              while (k > 0) {
                  y = y * 10 + k % 10;
                  k /= 10;
              }
              if (x + y == num) {
                  return true;
              }
          }
          return false;
      }
  };
  

• class Solution:
      def sumOfNumberAndReverse(self, num: int) -> bool:
          return any(k + int(str(k)[::-1]) == num for k in range(num + 1))
  
  

• func sumOfNumberAndReverse(num int) bool {
  	for x := 0; x <= num; x++ {
  		k, y := x, 0
  		for k > 0 {
  			y = y*10 + k%10
  			k /= 10
  		}
  		if x+y == num {
  			return true
  		}
  	}
  	return false
  }
  

• function sumOfNumberAndReverse(num: number): boolean {
      for (let i = 0; i <= num; i++) {
          if (i + Number([...(i + '')].reverse().join('')) === num) {
              return true;
          }
      }
      return false;
  }
  
  

• impl Solution {
      pub fn sum_of_number_and_reverse(num: i32) -> bool {
          for i in 0..=num {
              if
                  i +
                      ({
                          let mut t = i;
                          let mut j = 0;
                          while t > 0 {
                              j = j * 10 + (t % 10);
                              t /= 10;
                          }
                          j
                      }) == num
              {
                  return true;
              }
          }
          false
      }
  }
  
  

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