House Robber IV - Binary Search [JS]

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Description

Solution: Binary Search

Binary search for minimum maximum nums[i].
To check whether it is possible to take k non-adjacent houses with maximum score of nums[i],

  • Greedily take k houses with nums[i] <= max.
  • When we find nums[i] <= max, take the house and skip to nums[i + 2] (It is optimal to take a house earlier than later).

n = length of nums, m = max(nums[i])
Time Complexity: O(n log(m))
Space Complexity: O(1)

var minCapability = function(nums, k) {
  let min = nums[0], max = nums[0];
  let n = nums.length;
  for (let i = 0; i < n; i++) {
    min = Math.min(min, nums[i]);
    max = Math.max(max, nums[i]);
  }
  let low = min, high = max;
  while (low < high) {
    let mid = Math.floor((low + high) / 2);
    if (isEnough(mid)) high = mid;
    else low = mid + 1;
  }
  return low;
  
  function isEnough(max) { // greedily take k houses with nums[i] <= max
    let houses = 0;
    for (let i = 0; i < n; i++) {
      if (nums[i] <= max) {
        houses++;
        i++;
      }
      if (houses === k) return true;
    }
    return false;
  }
};
Location: Auckland, New Zealand

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