Algorithm #6: Array Part 2 - Minimum number of jumps to reach end

Questions 🔗︎

Given an array of integers where each element represents the max number of steps that can be made forward from that element. Write a function to return the minimum number of jumps to reach the end of the array (starting from the first element). If an element is 0, then we cannot move through that element. If we can’t reach the end, return -1. Examples:

Input: arr[] = {1, 3, 5, 8, 9, 2, 6, 7, 6, 8, 9} Output: 3 (1-> 3 -> 8 -> 9) Explanation: Jump from 1st element to 2nd element as there is only 1 step, now there are three options 5, 8 or 9. If 8 or 9 is chosen then the end node 9 can be reached. So 3 jumps are made.

Input: arr[] = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1} Output: 10 Explanation: In every step a jump is needed so the count of jumps is 10.

Solution 🔗︎

1 Method 1：Violent solution

//   i=1
// 1 0 0 0 0  1  4
// maxR 1 1
// step 1 2 1 0
// jump 2
function miniJumpsToEnd(arr) {
  const n = arr.length;
  if (n <= 1) return 0;
  if (arr[0] == 0) return -1;

  let maxReach = arr[0]; // 最大能走得到的索引
  let step = arr[0]; // 当前还剩几步
  let jumps = 1; // 总共走了多少步

  for (let index = 0; index < arr.length; index++) {
    if (index == arr.length - 1) {
      return jumps;
    }

    maxReach = Math.max(maxReach, index + arr[index]);

    step--;

    if (step == 0) {
      jumps++;

      // 不能继续往前走了 表示没有走到最后
      if (index >= maxReach) {
        return -1;
      }
      step = maxReach - index;
    }
  }
}

Time Complex 🔗︎

The violent solution is a bit non-violent ah, only traversed the time complexity is O(n) Record the current state by three variables, the space complexity is O(1)