Problem

Given an integer num, return a string of its base 7 representation.

Examples

Example 1

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Input: num = 100
Output: "202"

Example 2

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Input: num = -7
Output: "-10"

Constraints

  • -10^7 <= num <= 10^7

Solution

Method 1 – Repeated Division (Base Conversion)

Intuition

To convert a decimal number to base 7, repeatedly divide the number by 7 and collect the remainders. The remainders, read in reverse order, form the base 7 representation. Handle negative numbers by converting the absolute value and adding a minus sign if needed.

Approach

  1. If num is 0, return "0".
  2. Record if num is negative; work with its absolute value.
  3. While num is greater than 0:
  • Divide num by 7, store the remainder.
  • Update num to the quotient.
  • Append the remainder to a result list or string.
  1. Reverse the collected digits.
  2. Add a minus sign if the original number was negative.
  3. Return the result as a string.

Code

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class Solution {
public:
   string convertToBase7(int num) {
      if (num == 0) return "0";
      bool neg = num < 0;
      num = abs(num);
      string ans;
      while (num > 0) {
        ans += to_string(num % 7);
        num /= 7;
      }
      if (neg) ans += '-';
      reverse(ans.begin(), ans.end());
      return ans;
   }
};
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func convertToBase7(num int) string {
   if num == 0 {
      return "0"
   }
   neg := num < 0
   if neg {
      num = -num
   }
   ans := ""
   for num > 0 {
      ans = string('0'+num%7) + ans
      num /= 7
   }
   if neg {
      ans = "-" + ans
   }
   return ans
}
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class Solution {
   public String convertToBase7(int num) {
      if (num == 0) return "0";
      boolean neg = num < 0;
      num = Math.abs(num);
      StringBuilder ans = new StringBuilder();
      while (num > 0) {
        ans.append(num % 7);
        num /= 7;
      }
      if (neg) ans.append('-');
      return ans.reverse().toString();
   }
}
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class Solution {
   fun convertToBase7(num: Int): String {
      if (num == 0) return "0"
      var n = Math.abs(num)
      val neg = num < 0
      val ans = StringBuilder()
      while (n > 0) {
        ans.append(n % 7)
        n /= 7
      }
      if (neg) ans.append('-')
      return ans.reverse().toString()
   }
}
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class Solution:
   def convertToBase7(self, num: int) -> str:
      if num == 0:
        return "0"
      neg = num < 0
      n = abs(num)
      ans = ""
      while n > 0:
        ans = str(n % 7) + ans
        n //= 7
      return "-" + ans if neg else ans
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impl Solution {
   pub fn convert_to_base7(num: i32) -> String {
      if num == 0 {
        return "0".to_string();
      }
      let mut n = num.abs();
      let mut ans = String::new();
      while n > 0 {
        ans.insert(0, char::from_digit((n % 7) as u32, 10).unwrap());
        n /= 7;
      }
      if num < 0 {
        ans.insert(0, '-');
      }
      ans
   }
}

Complexity

  • ⏰ Time complexity: O(log₇|num|) — Each division reduces the number by a factor of 7.
  • 🧺 Space complexity: O(log₇|num|) — For storing the digits of the result.