List all cases in which M of the numbers from 1 to N are selected without overlapping. If the order is different, even the same value is treated as different cases. (1 <= M <= N)

backtracking-algorithm-iterative-solution.ts

const [N,M] = [5, 2];
const contextStack: number[][] = [];
const stack = Array.from({length: N}, (_,i) => i + 1);
const result: number[] = [];
let target: number;
let isEnd = false;
while (1) {
  // When one case is completed
  if (result.length === M) {
    console.log(result.join(' '));
    result.pop();
    continue;
  }

  // If run out of stack, go back to the previous context.
  while (!stack.length) {
    // If don't have both the remaining stack and the context stack, it's over.
    if (!stack.length && !contextStack.length) {
      isEnd = true;
      break;
    }
    
    stack.push(...contextStack.pop() as number[]); 
    result.pop();
  }
  if (isEnd) break;

  // The currently selected number.
  target = stack.shift();

  // Pass the number that was already chosen.
  if (!result.includes(target)) {
    result.push(target);

    // Save context so that you can go back if you don't choose all M of Numbers.
    if (result.length !== M) {
      contextStack.push(stack.splice(0));
      // Stack reset.
      stack.push(...Array.from({length: N}, (_,i) => i + 1));
    }
  }
}

// result console.log
// 1 2
// 1 3
// 1 4
// 1 5
// 2 1
// 2 3
// 2 4
// 2 5
// 3 1
// 3 2
// 3 4
// 3 5
// 4 1
// 4 2
// 4 3
// 4 5
// 5 1
// 5 2
// 5 3
// 5 4