Leetcode 1564

The idea of this solution is:

We sort boxes.
Then, we make an array called minimums, and put the minimum values of warehouse from left to right into the array. If you don’t understand, look at the following images:
Then we can just look from len(minimums) - 1 to 0 and check whether minimums[i] >= boxes[boxCounter] we can go to the next box, other wise we continue going through the loop until minimums[i] >= boxes[boxCounter].
Then we can return boxCounter because boxCounter = the number of boxes we have boxes we have fitted in.

func maxBoxesInWarehouse(boxes []int, warehouse []int) int {
    boxCounter := 0
    sort.Ints(boxes)
    minimums := []int{warehouse[0]}

    for i := 1; i < len(warehouse); i++ {
        if minimums[i - 1] > warehouse[i] {
            minimums = append(minimums, warehouse[i])
        } else {
            minimums = append(minimums, minimums[i - 1])
        }
    }

    for i := len(minimums) - 1; i >= 0; i-- {
        if boxCounter == len(boxes) {
            break
        }
        if minimums[i] >= boxes[boxCounter] {
            boxCounter++
        }
    }

    return boxCounter
}