/* Given a string of characters. Find the number of pairs of indices i and j (i <= j), between which there are no repeating characters. For the string "aba" the answer is 5: [0, 0] ("a") [0, 1] ("ab") [1, 1] ("b") [1, 2] ("ba") [2, 2] ("a") */

To find the number of index pairs (i, j) for which the substring s[i:j+1] contains no repeating characters, you can use the "sliding window" method (two pointers).

Idea:

• Use two pointers: start and end, which will denote the current considered interval.
• Expand end while the characters are unique.
• When a repeat is encountered, move start, removing characters from the set.
• For each position of end, count the number of substrings without repeats that end at end.

Example in Go:

func countUniqueSubstrings(s string) int {
    seen := make(map[rune]int)
    start := 0
    count := 0
    for end, ch := range s {
        if pos, ok := seen[ch]; ok && pos >= start {
            start = pos + 1
        }
        seen[ch] = end
        count += end - start + 1
    }
    return count
}

Here, count accumulates the number of unique substrings without repeating characters.