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This approach uses a simple greedy algorithm to find the first occurrence of the digit '6' and replaces it with '9'. This ensures that the number is maximized by making a change that increases the most significant portion of the number.
Time Complexity: O(n), where n is the number of digits in num.
Space Complexity: O(n) for storing the string representation of num.
1public class Solution {
2 public int maximum69Number (int num) {
3 String numStr = Integer.toString(num);
4 char[] numArray = numStr.toCharArray();
5 for (int i = 0; i < numArray.length; i++) {
6 if (numArray[i] == '6') {
7 numArray[i] = '9';
8 break;
9 }
10 }
11 return Integer.parseInt(new String(numArray));
12 }
13
14 public static void main(String[] args) {
15 Solution sol = new Solution();
16 System.out.println(sol.maximum69Number(9669));
17 }
18}The Java solution leverages toString to convert the number into a character array for traversal, where it changes the first '6' to '9'. After the loop finishes, the new string representation is parsed back to an integer.
This approach works directly with the number without converting it to a string. We find the first '6' from the leftmost side by using basic mathematical operations (division and modulo). We can calculate the position and change the digit using arithmetic transformations to make this efficient.
Time Complexity: O(1), as we have at most 4 digits to check.
Space Complexity: O(1), using constant space for computations.
1using namespace std;
int maximum69Number (int num) {
int mod = 10000;
for (int i = 4; i >= 0; i--) {
if ((num / mod) % 10 == 6) {
num += 3 * mod;
break;
}
mod /= 10;
}
return num;
}
int main() {
int num = 9669;
cout << maximum69Number(num);
return 0;
}The C++ solution uses integer division to figure out the value of individual digits and alters the first '6' to '9' using arithmetic.