Easy aproach with explanation
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```javascript []\n/**\n * @param {string} s1\n * @param {string} s2\n * @return {string[]}\n */var uncommonFromSentences = function(s1, s2) {\n    // Create a new map to store the frequency of each word.\n    let m = new Map()\n    \n    // Split the first sentence \'s1\' into an array of words.\n    s1 = s1.split(\' \')\n\n    // Iterate through each word in the first sentence (s1).\n    for (let i = 0; i < s1.length; i++) {\n        // If the word already exists in the map, update its frequency (increase count by 1).\n        if (m.has(s1[i])) {\n            // m.get(s1[i]) returns the current count of the word, and we add 1 to it.\n            m.set(s1[i], m.get(s1[i]) + 1)\n        } else {\n            // If the word doesn\'t exist in the map, add it with an initial frequency of 1.\n            m.set(s1[i], 1)\n        }\n    }\n\n    // Split the second sentence \'s2\' into an array of words.\n    s2 = s2.split(\' \')\n\n    // Iterate through each word in the second sentence (s2).\n    for (let i = 0; i < s2.length; i++) {\n        // If the word already exists in the map (from s1 or s2), update its frequency.\n        if (m.has(s2[i])) {\n            // m.get(s2[i]) returns the current count of the word, and we add 1 to it.\n            m.set(s2[i], m.get(s2[i]) + 1)\n        } else {\n            // If the word doesn\'t exist in the map, add it with an initial frequency of 1.\n            m.set(s2[i], 1)\n        }\n    }\n\n    // Create an empty array to store the result (uncommon words).\n    let result = []\n\n    // Iterate through the map to check the frequency of each word.\n    for (let [key, val] of m) {\n        // If the frequency of the word is exactly 1, it means the word is uncommon.\n        if (val === 1) {\n            // Add the uncommon word to the result array.\n            result.push(key)\n        }\n    }\n\n    // Return the array of uncommon words.\n    return result\n};\n
['Hash Table', 'String', 'Counting', 'TypeScript', 'JavaScript']