Substring with Concatenation of All Words

All words have equal length w, so scan the string in w separate alignments, not one character-by-character window. Build need, the frequency map of the words. For each offset from 0 to w-1, move right in steps of w, read the next chunk, and update a seen map and word count. If a chunk is not required, clear the window and restart after it. If a word becomes overrepresented, advance left by w chunks until counts are valid again. When the window contains exactly words.length chunks, record left, then remove its first word to continue searching overlapping matches. This respects duplicate words. Let n be string length and m number of words: time is O(n) map operations plus substring extraction costs, and space is O(m) distinct-word state. Ask what to do if word lengths differ; this method relies on equality.