# My solution used the same pattern as String Anagrams (Problem Challenge 2)

const find_word_concatenation = function(str, words) {
result_indices = [];
// this is the same as find starting indices of all anagrams
// this just needs to jump windowEnd, windowStart by 3 instead of 1

// create a frequencyMap from words
const frequencyMap = {};
for (i=0 ; i<words.length ; i++) {
const word = words[i]; // cat
if (word in frequencyMap) {
frequencyMap[word] += 1;
} else {
frequencyMap[word] = 1;
}
}

const wordLength = words[0].length;
let windowStart = 0;
let matched = 0;
// form a sliding window in a stepwise fashion
for (windowEnd = 0 ; windowEnd < str.length ; windowEnd += wordLength) {
// check if current word in the window is an anagram of the words
const rightWord = str.substring(windowEnd, windowEnd + wordLength);
if (rightWord in frequencyMap) {
frequencyMap[rightWord] -= 1;
if (frequencyMap[rightWord] >= 0) {
matched += 1;
}

// if true, insert the starting index to the result
if (matched === words.length) {
result_indices.push(windowStart);
}
}

// if window length === words length (char length), shrink the window
if (windowEnd - windowStart + wordLength === words.length * wordLength) {
// shrink the window from the left, and adjust values in frequencyMap and matched
const leftWord = str.substring(windowStart, windowStart + wordLength);
if (leftWord in frequencyMap) {
if (frequencyMap[leftWord] >= 0) {
matched -= 1;
}
frequencyMap[leftWord] += 1;
}
windowStart += wordLength;
}
}

// return result
return result_indices;
}

@Minjun_Youn