Arrays and Iteration

Find Maximum

Scan the array once, keeping the largest value seen so far. The replay highlights when a candidate replaces the running maximum.

Algorithm

Basic Implementation

basic.ts 
const arr: number[] = [3, 1, 4, 1, 5, 9, 2, 6];
let best: number = arr[0];
for (let i = 1; i < arr.length; i++) {
  if (arr[i] > best) {
    best = arr[i];
  }
}
console.log(best);

Complexity

  • Time: O(n)
  • Space: O(1)

Implementation notes

  • TypeScript declares the input as const arr: number[] and the running value as let best: number, so the checked source scans finite JavaScript Number values through TypeScript's number type.
  • The implementation uses an indexed for loop rather than reduce, so the state stays in the local best binding seeded from arr[0], with arr[i] as the current value.
  • Each arr[i] > best comparison is numeric. When true, best = arr[i] rebinds the local number; the array itself is never mutated.
  • The replay shows replacements at i=2 (4), i=4 (5), and i=5 (9), while values 1, 1, 2, and 6 leave best unchanged.
  • console.log(best) prints 9. Aside from the initial number[], the scan only updates scalar bindings and creates no lesson-visible heap allocation.
execution replay The checked-in replay follows the language-neutral state table for `array-find-max`.
cross-language comparison This TypeScript DSA version keeps the same data and final output as every other DSA book in this wave.