Connected to surface test questions

An interesting interview question , Record
An array [1,0,1], Think of him as a bar chart , Then you can get water 1,

An array [2,0,3], Think of him as a bar chart , Then you can get water 2.
Find a statistical method of water saving .

One . Violence law

function demo(arr=[]) {
    
	if(Array.isArray(arr) && arr.length > 2) {
    
		let sum = 0
		let len = arr.length
		for(let i = 1; i < len - 1; i ++) {
    
			let cur = arr[i]
			let leftMax = 0
			let rightMax = 0
			for(let j = i; j < len; j ++) {
    
				rightMax = Math.max(arr[j], rightMax)
			}
			for(let j = i; j >= 0; j --) {
    
				leftMax = Math.max(arr[j], leftMax)
			}
			sum += (Math.min(leftMax, rightMax) - cur)
		}
		return sum
	}else {
    
		return 0
	}
}

Two . Violence act simplified

function demo(arr=[]) {
    
	if(Array.isArray(arr) && arr.length > 2) {
    
		let sum = 0
		let len = arr.length
		for(let i = 1; i < len - 1; i ++) {
    
			let cur = arr[i]
			let leftMax = Math.max(...arr.slice(0, i + 1))
			let rightMax = Math.max(...arr.slice(i))
			sum += (Math.min(rightMax, leftMax) - cur)
		}
		return sum
	}else {
    
		return 0
	}
}

3、 ... and . Double finger needling

function demo(arr = []) {
    
	if (Array.isArray(arr) && arr.length > 2) {
    
		let [sum, len] = [0, arr.length]
		let [leftIndex, rightIndex] = [0, len - 1]
		let [leftMax, rightMax] = [arr[leftIndex], arr[rightIndex]]
		while (leftIndex < rightIndex) {
    
			leftMax = Math.max(leftMax, arr[leftIndex])
			rightMax = Math.max(rightMax, arr[rightIndex])
			if (leftMax < rightMax) {
    
				sum += (leftMax - arr[leftIndex])
				leftIndex++
			} else {
    
				sum += (rightMax - arr[rightIndex])
				rightIndex--
			}
		}
		return sum
	} else {
    
		return 0
	}
}