Get Equal Substrings Within Budget

The strings are aligned, so changing s[i] to t[i] costs abs(code(s[i])-code(t[i])). Maintain a window [left,right] and its total conversion cost. Expand right by adding its aligned cost; while total exceeds maxCost, subtract the left cost and advance left. After that loop, the window is valid, so update the best length. This works because all individual costs are nonnegative: expanding cannot reduce cost, and shrinking cannot increase it, which makes feasibility monotonic. Test zero budget, identical strings, one expensive character, the best window at the beginning or end, and equal-length validation; if strings can differ in length, clarify the intended alignment first. Each pointer only moves forward, so time is O(n) and extra space is O(1). Avoid recomputing each window's full cost, which produces O(n squared) work.