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This approach constructs the string by always choosing the majority character until we need to switch to prevent three consecutive characters.
Steps:
Time Complexity: O(a + b) because each character is processed once.
Space Complexity: O(a + b) for the result string.
1#include <stdio.h>
2#include <string.h>
3
4char* strWithout3a3b(int a, int b) {
5 static char result[201];
6 int index = 0;
7 while (a > 0 || b > 0) {
8 int useA = 0;
9 if (index > 1 && result[index-1] == result[index-2]) {
10 useA = result[index-1] == 'b';
11 } else {
12 useA = a >= b;
13 }
14 if (useA) {
15 result[index++] = 'a';
16 a--;
17 } else {
18 result[index++] = 'b';
19 b--;
20 }
21 }
22 result[index] = '\0';
23 return result;
24}
25
26int main() {
27 int a = 4, b = 1;
28 printf("%s\n", strWithout3a3b(a, b));
29 return 0;
30}This C solution uses a static buffer result which is filled character by character based on the remaining count of 'a's and 'b's. The decision on whether to append 'a' or 'b' depends on the count of the last two characters, as well as the overall remaining count of 'a' and 'b'.
This approach leverages an interleaving strategy to distribute characters evenly and avoid consecutive identical sequences.
Steps:
Time Complexity: O(a + b) due to iteration through characters.
Space Complexity: O(a + b) used for the output string.
1#include <string>
std::string strWithout3a3b(int a, int b) {
std::string result;
while (a > 0 || b > 0) {
if (a > b) {
if (a > 0) { result += 'a'; a--; }
if (a > 0) { result += 'a'; a--; }
if (b > 0) { result += 'b'; b--; }
} else {
if (b > 0) { result += 'b'; b--; }
if (b > 0) { result += 'b'; b--; }
if (a > 0) { result += 'a'; a--; }
}
}
return result;
}
int main() {
std::cout << strWithout3a3b(4, 1) << std::endl;
return 0;
}This C++ solution uses a simple strategy to build the string iteratively by adding the characters according to their frequency and ensuring no 'aaa' or 'bbb' sequences arise due to the controlled interleaving.