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Lexographically smallest after a swap (#99)
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LeetcodeProblems/Algorithms/easy/Lexographic_smallest_after_swap.js
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/* | ||
3216. Lexicographically Smallest String After a Swap | ||
https://leetcode.com/problems/lexicographically-smallest-string-after-a-swap/description/ | ||
Problem: | ||
Given a string s containing only digits, return the lexicographically smallest string | ||
that can be obtained after swapping adjacent digits in s with the same parity at most once. | ||
Digits have the same parity if both are odd or both are even. For example, 5 and 9, as well as 2 and 4, | ||
have the same parity, while 6 and 9 do not. | ||
Example 1: | ||
Input: s = "45320" | ||
Output: "43520" | ||
Explanation: | ||
s[1] == '5' and s[2] == '3' both have the same parity, and swapping them results in the lexicographically smallest string. | ||
Example 2: | ||
Input: s = "001" | ||
Output: "001" | ||
Explanation: | ||
There is no need to perform a swap because s is already the lexicographically smallest. | ||
Constraints: | ||
2 <= s.length <= 100 | ||
s consists only of digits. | ||
*/ | ||
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/* | ||
Approach: | ||
Checking if the present digit is greater than the next digit, if yes then checking the parity of the digit. | ||
if both have the same parity swap the number. since this operation can be done at max once, break the loop after the first swap. | ||
return the updated number | ||
What is parity of a number? | ||
Parity: the property of an integer of whether it is even or odd | ||
how to check the parity of the number: | ||
- using the & operation on the last bit, | ||
- if (last bit)&1 == 1, means the last bit was 1. means Odd (ex: 3 has a bit representation of 11) | ||
- if (last bit)&1 == 0, means last bit was 0. means Even number ( ex: 2 has a bit representation of 10) | ||
*/ | ||
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/** | ||
* @param {string} s | ||
* @return {string} | ||
*/ | ||
var getSmallestString = function(s) { | ||
let arr = s.split("").map(Number); | ||
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const getParity = (num) => { | ||
if(num&1 === 0) return "even"; | ||
else return "odd"; | ||
}; | ||
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for(let i = 0; i< s.length - 1; i++) { | ||
if(arr[i] > arr[i+1] && getParity(arr[i]) === getParity(arr[i + 1])) { | ||
let tmp = arr[i+1]; | ||
arr[i+1] = arr[i]; | ||
arr[i] = tmp; | ||
break; | ||
} | ||
} | ||
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return arr.join(""); | ||
}; | ||
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module.exports.getSmallestString = getSmallestString; |
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LeetcodeProblemsTests/Algorithms/easy/Lexographic_smallest_after_swap.js
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const assert = require("assert"); | ||
const bitReverseToMakeNumberEqual = require("../../../LeetcodeProblems/Algorithms/easy/Lexographic_smallest_after_swap").getSmallestString; | ||
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function test() { | ||
assert.deepEqual( | ||
getSmallestString("45320"), | ||
"43520" | ||
); | ||
assert.deepEqual( | ||
getSmallestString("001"), | ||
"001" | ||
); | ||
); | ||
} | ||
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module.exports.test = test; |
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