Given two integers A and B return string containing letters “a” and “b” with no three consecutive...
-2
Asked in coding challenge.
Example:
A = 5 and B = 3 return "aabaabab" or "abbaabaa" both are correct answers
A = 1 and B = 4 return "bbabb"
constraints : ignore cases where A = 0 and B >= 3 vice-versa
string algorithm data-structures greedy
asked Nov 13 at 6:18
Shailesh Shekhawat
135113
add a comment |
-2
Asked in coding challenge.
Example:
A = 5 and B = 3 return "aabaabab" or "abbaabaa" both are correct answers
A = 1 and B = 4 return "bbabb"
constraints : ignore cases where A = 0 and B >= 3 vice-versa
string algorithm data-structures greedy
asked Nov 13 at 6:18
Shailesh Shekhawat
135113
Seems there is something missing in the question, or is this it?
– Surt
Nov 13 at 6:20
1
So, what is your problem?
– Andreas
Nov 13 at 6:28
This is it. We need to write a function that will return string like above
– Shailesh Shekhawat
Nov 13 at 6:30
1
Compareaandbvalues. Let's say ifais larger, then create a string with onlya's and then insertbafter every 2a's. If there are stillb's left, then insertbafterbthat was attached.
– vivek_23
Nov 13 at 8:04
if there are stillb's left, attach them at the beginning and end of the string.
– vivek_23
Nov 13 at 8:34
add a comment |
-2
-2
-2
Asked in coding challenge.
Example:
A = 5 and B = 3 return "aabaabab" or "abbaabaa" both are correct answers
A = 1 and B = 4 return "bbabb"
constraints : ignore cases where A = 0 and B >= 3 vice-versa
string algorithm data-structures greedy
asked Nov 13 at 6:18
Shailesh Shekhawat
135113
Asked in coding challenge.
Example:
A = 5 and B = 3 return "aabaabab" or "abbaabaa" both are correct answers
A = 1 and B = 4 return "bbabb"
constraints : ignore cases where A = 0 and B >= 3 vice-versa
string algorithm data-structures greedy
string algorithm data-structures greedy
asked Nov 13 at 6:18
Shailesh Shekhawat
135113
asked Nov 13 at 6:18
Shailesh Shekhawat
135113
asked Nov 13 at 6:18
Shailesh Shekhawat
135113
asked Nov 13 at 6:18
Shailesh Shekhawat
135113
asked Nov 13 at 6:18
Shailesh Shekhawat
135113
135113
Seems there is something missing in the question, or is this it?
– Surt
Nov 13 at 6:20
1
So, what is your problem?
– Andreas
Nov 13 at 6:28
This is it. We need to write a function that will return string like above
– Shailesh Shekhawat
Nov 13 at 6:30
1
Compareaandbvalues. Let's say ifais larger, then create a string with onlya's and then insertbafter every 2a's. If there are stillb's left, then insertbafterbthat was attached.
– vivek_23
Nov 13 at 8:04
if there are stillb's left, attach them at the beginning and end of the string.
– vivek_23
Nov 13 at 8:34
add a comment |
Seems there is something missing in the question, or is this it?
– Surt
Nov 13 at 6:20
1
So, what is your problem?
– Andreas
Nov 13 at 6:28
This is it. We need to write a function that will return string like above
– Shailesh Shekhawat
Nov 13 at 6:30
1
Compareaandbvalues. Let's say ifais larger, then create a string with onlya's and then insertbafter every 2a's. If there are stillb's left, then insertbafterbthat was attached.
– vivek_23
Nov 13 at 8:04
if there are stillb's left, attach them at the beginning and end of the string.
– vivek_23
Nov 13 at 8:34
Seems there is something missing in the question, or is this it?
– Surt
Nov 13 at 6:20
Seems there is something missing in the question, or is this it?
– Surt
Nov 13 at 6:20
1
1
So, what is your problem?
– Andreas
Nov 13 at 6:28
So, what is your problem?
– Andreas
Nov 13 at 6:28
This is it. We need to write a function that will return string like above
– Shailesh Shekhawat
Nov 13 at 6:30
This is it. We need to write a function that will return string like above
– Shailesh Shekhawat
Nov 13 at 6:30
1
1
Compare
a and b values. Let's say if a is larger, then create a string with only a's and then insert b after every 2 a's. If there are still b's left, then insert b after b that was attached.– vivek_23
Nov 13 at 8:04
Compare
a and b values. Let's say if a is larger, then create a string with only a's and then insert b after every 2 a's. If there are still b's left, then insert b after b that was attached.– vivek_23
Nov 13 at 8:04
if there are still
b's left, attach them at the beginning and end of the string.– vivek_23
Nov 13 at 8:34
if there are still
b's left, attach them at the beginning and end of the string.– vivek_23
Nov 13 at 8:34
add a comment |
2 Answers
2
active
oldest
votes
2
As per above problem my JAVA solution is:
public static String solution(int A, int B) {
String biggerString, smallerString;
int biggerCount, smallerCount;
if (A >= B) {
biggerString = "a";
smallerString = "b";
biggerCount = A;
smallerCount = B;
} else {
biggerString = "b";
smallerString = "a";
biggerCount = B;
smallerCount = A;
}
// A > 2B + 2
if (biggerCount > 2 * smallerCount + 2) {
return "No Soulution";
}
StringBuilder returnString = new StringBuilder();
do {
if (biggerCount == smallerCount + 1) {
returnString.append(biggerString);
biggerCount--;
} else if (biggerCount == smallerCount) {
returnString.append(biggerString);
biggerCount--;
returnString.append(smallerString);
smallerCount--;
} else {
returnString.append(biggerString);
returnString.append(biggerString);
biggerCount -= 2;
if (smallerCount > 0) {
returnString.append(smallerString);
smallerCount -= 1;
}
}
} while (biggerCount > 0);
return returnString.toString();
}
answered Nov 22 at 17:41
Amrith Raj Herle
1268
add a comment |
1
Assume A is greater than B, otherwise swap them (keep the track of which belongs to which letter).
If A > 2B + 2, there is no solution, as you don't have enough letters b to fill the gaps between a's.
Keep inserting 2 a's and 1 b's until there is equal number of them or there is no more b's. After that place 1 of each consecutively.
At the end you'll be left with no more letters, or 1 more letter a. Place it at the end. You're done.
answered Nov 13 at 9:49
hazeiio
1796
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
As per above problem my JAVA solution is:
public static String solution(int A, int B) {
String biggerString, smallerString;
int biggerCount, smallerCount;
if (A >= B) {
biggerString = "a";
smallerString = "b";
biggerCount = A;
smallerCount = B;
} else {
biggerString = "b";
smallerString = "a";
biggerCount = B;
smallerCount = A;
}
// A > 2B + 2
if (biggerCount > 2 * smallerCount + 2) {
return "No Soulution";
}
StringBuilder returnString = new StringBuilder();
do {
if (biggerCount == smallerCount + 1) {
returnString.append(biggerString);
biggerCount--;
} else if (biggerCount == smallerCount) {
returnString.append(biggerString);
biggerCount--;
returnString.append(smallerString);
smallerCount--;
} else {
returnString.append(biggerString);
returnString.append(biggerString);
biggerCount -= 2;
if (smallerCount > 0) {
returnString.append(smallerString);
smallerCount -= 1;
}
}
} while (biggerCount > 0);
return returnString.toString();
}
answered Nov 22 at 17:41
Amrith Raj Herle
1268
add a comment |
2
As per above problem my JAVA solution is:
public static String solution(int A, int B) {
String biggerString, smallerString;
int biggerCount, smallerCount;
if (A >= B) {
biggerString = "a";
smallerString = "b";
biggerCount = A;
smallerCount = B;
} else {
biggerString = "b";
smallerString = "a";
biggerCount = B;
smallerCount = A;
}
// A > 2B + 2
if (biggerCount > 2 * smallerCount + 2) {
return "No Soulution";
}
StringBuilder returnString = new StringBuilder();
do {
if (biggerCount == smallerCount + 1) {
returnString.append(biggerString);
biggerCount--;
} else if (biggerCount == smallerCount) {
returnString.append(biggerString);
biggerCount--;
returnString.append(smallerString);
smallerCount--;
} else {
returnString.append(biggerString);
returnString.append(biggerString);
biggerCount -= 2;
if (smallerCount > 0) {
returnString.append(smallerString);
smallerCount -= 1;
}
}
} while (biggerCount > 0);
return returnString.toString();
}
answered Nov 22 at 17:41
Amrith Raj Herle
1268
add a comment |
2
2
2
As per above problem my JAVA solution is:
public static String solution(int A, int B) {
String biggerString, smallerString;
int biggerCount, smallerCount;
if (A >= B) {
biggerString = "a";
smallerString = "b";
biggerCount = A;
smallerCount = B;
} else {
biggerString = "b";
smallerString = "a";
biggerCount = B;
smallerCount = A;
}
// A > 2B + 2
if (biggerCount > 2 * smallerCount + 2) {
return "No Soulution";
}
StringBuilder returnString = new StringBuilder();
do {
if (biggerCount == smallerCount + 1) {
returnString.append(biggerString);
biggerCount--;
} else if (biggerCount == smallerCount) {
returnString.append(biggerString);
biggerCount--;
returnString.append(smallerString);
smallerCount--;
} else {
returnString.append(biggerString);
returnString.append(biggerString);
biggerCount -= 2;
if (smallerCount > 0) {
returnString.append(smallerString);
smallerCount -= 1;
}
}
} while (biggerCount > 0);
return returnString.toString();
}
answered Nov 22 at 17:41
Amrith Raj Herle
1268
As per above problem my JAVA solution is:
public static String solution(int A, int B) {
String biggerString, smallerString;
int biggerCount, smallerCount;
if (A >= B) {
biggerString = "a";
smallerString = "b";
biggerCount = A;
smallerCount = B;
} else {
biggerString = "b";
smallerString = "a";
biggerCount = B;
smallerCount = A;
}
// A > 2B + 2
if (biggerCount > 2 * smallerCount + 2) {
return "No Soulution";
}
StringBuilder returnString = new StringBuilder();
do {
if (biggerCount == smallerCount + 1) {
returnString.append(biggerString);
biggerCount--;
} else if (biggerCount == smallerCount) {
returnString.append(biggerString);
biggerCount--;
returnString.append(smallerString);
smallerCount--;
} else {
returnString.append(biggerString);
returnString.append(biggerString);
biggerCount -= 2;
if (smallerCount > 0) {
returnString.append(smallerString);
smallerCount -= 1;
}
}
} while (biggerCount > 0);
return returnString.toString();
}
answered Nov 22 at 17:41
Amrith Raj Herle
1268
answered Nov 22 at 17:41
Amrith Raj Herle
1268
answered Nov 22 at 17:41
Amrith Raj Herle
1268
answered Nov 22 at 17:41
Amrith Raj Herle
1268
1268
add a comment |
add a comment |
1
Assume A is greater than B, otherwise swap them (keep the track of which belongs to which letter).
If A > 2B + 2, there is no solution, as you don't have enough letters b to fill the gaps between a's.
Keep inserting 2 a's and 1 b's until there is equal number of them or there is no more b's. After that place 1 of each consecutively.
At the end you'll be left with no more letters, or 1 more letter a. Place it at the end. You're done.
answered Nov 13 at 9:49
hazeiio
1796
add a comment |
1
Assume A is greater than B, otherwise swap them (keep the track of which belongs to which letter).
If A > 2B + 2, there is no solution, as you don't have enough letters b to fill the gaps between a's.
Keep inserting 2 a's and 1 b's until there is equal number of them or there is no more b's. After that place 1 of each consecutively.
At the end you'll be left with no more letters, or 1 more letter a. Place it at the end. You're done.
answered Nov 13 at 9:49
hazeiio
1796
add a comment |
1
1
1
Assume A is greater than B, otherwise swap them (keep the track of which belongs to which letter).
If A > 2B + 2, there is no solution, as you don't have enough letters b to fill the gaps between a's.
Keep inserting 2 a's and 1 b's until there is equal number of them or there is no more b's. After that place 1 of each consecutively.
At the end you'll be left with no more letters, or 1 more letter a. Place it at the end. You're done.
answered Nov 13 at 9:49
hazeiio
1796
Assume A is greater than B, otherwise swap them (keep the track of which belongs to which letter).
If A > 2B + 2, there is no solution, as you don't have enough letters b to fill the gaps between a's.
Keep inserting 2 a's and 1 b's until there is equal number of them or there is no more b's. After that place 1 of each consecutively.
At the end you'll be left with no more letters, or 1 more letter a. Place it at the end. You're done.
answered Nov 13 at 9:49
hazeiio
1796
answered Nov 13 at 9:49
hazeiio
1796
answered Nov 13 at 9:49
hazeiio
1796
answered Nov 13 at 9:49
hazeiio
1796
1796
add a comment |
add a comment |
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Seems there is something missing in the question, or is this it?
– Surt
Nov 13 at 6:20
1
So, what is your problem?
– Andreas
Nov 13 at 6:28
This is it. We need to write a function that will return string like above
– Shailesh Shekhawat
Nov 13 at 6:30
1
Compare
aandbvalues. Let's say ifais larger, then create a string with onlya's and then insertbafter every 2a's. If there are stillb's left, then insertbafterbthat was attached.– vivek_23
Nov 13 at 8:04
if there are still
b's left, attach them at the beginning and end of the string.– vivek_23
Nov 13 at 8:34