LeetCode: Last Substring in Lexicographical Order

// https://code.dennyzhang.com/last-substring-in-lexicographical-order
// Basic Ideas: greedy + two pointer
//
//   The ending point of the result is the last character.
//   Need to define where is the starting point.
//   Apparently we know the value of the starting point.
//   It must be the biggest character in the string.
//
//   Use two slow and fast pointer:
//       - slow: starting point of the result
//       - fast: starting point of the candidate
//
//   The key part is slow pointer won't need to move back
//
//   If candidate of faster pointer is better than the slow pointer, 
//      just need to move the slow pointer to faster point.
//
//   No need to examining starting in between slow and fast characters!
//
//   cabcabcb
//   .  .  .
//
//   
// Complexity: Time O(n), Space O(1)
func max(x, y int) int {
    if x>y {
        return x
    } else {
        return y
    }
}

func lastSubstring(s string) string {
    i, j := 0, 1
    for k:=0; k+j<len(s); {
        if s[i+k] == s[j+k] {
            k++
            continue
        }
        if s[i+k] > s[j+k] {
            j = j+k+1
        } else {
            // Think: why not simply i=j
            i = max(i+k+1, j)
            j = i+1
        }
        k = 0
    }
    return s[i:]
}