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| 1 | +// Alternative arrange the two given strings in one string in O(n) time complexity. |
| 2 | + |
| 3 | +// Problem Source & Explanation: https://www.geeksforgeeks.org/alternatively-merge-two-strings-in-java/ |
| 4 | + |
| 5 | +/** |
| 6 | + * Alternative arrange the two given strings in one string in O(n) time complexity. |
| 7 | + * @param {String} str1 first input string |
| 8 | + * @param {String} str2 second input string |
| 9 | + * @returns `String` return one alternative arrange string. |
| 10 | + */ |
| 11 | +const AlternativeStringArrange = (str1, str2) => { |
| 12 | + // firstly, check that both inputs are strings. |
| 13 | + if (typeof str1 !== 'string' || typeof str2 !== 'string') { |
| 14 | + return 'Not string(s)' |
| 15 | + } |
| 16 | + |
| 17 | + // output string vlaue. |
| 18 | + let outStr = '' |
| 19 | + |
| 20 | + // get first string length. |
| 21 | + const firstStringLength = str1.length |
| 22 | + // get second string length. |
| 23 | + const secondStringLength = str2.length |
| 24 | + // absolute length for oparetion. |
| 25 | + const absLenght = firstStringLength > secondStringLength ? firstStringLength : secondStringLength |
| 26 | + |
| 27 | + // Iterate the character count until the absolute count is reached. |
| 28 | + for (let charCount = 0; charCount < absLenght; charCount++) { |
| 29 | + // If firstStringLength is lesser than the charCount it means they are able to re-arange. |
| 30 | + if (charCount < firstStringLength) { |
| 31 | + outStr += str1[charCount] |
| 32 | + } |
| 33 | + |
| 34 | + // If secondStringLength is lesser than the charCount it means they are able to re-arange. |
| 35 | + if (charCount < secondStringLength) { |
| 36 | + outStr += str2[charCount] |
| 37 | + } |
| 38 | + } |
| 39 | + |
| 40 | + // return the output string. |
| 41 | + return outStr |
| 42 | +} |
| 43 | + |
| 44 | +module.exports = AlternativeStringArrange |
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