# 10.07.2023 [2272. Substring With Largest Variance]

Max diff between count `s[i]` and count `s[j]` in all substrings of `s`

10.07.2023

2272. Substring With Largest Variance hard
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Problem TLDR

Max diff between count s[i] and count s[j] in all substrings of s

Intuition

The first idea is to simplify the task by considering only two chars, iterating over all alphabet combinations.
Second idea is how to solve this problem for binary string in O(n): abaabbb → abbb.
We split this problem: find the largest subarray for a with the smallest count of b, and reverse the problem – largest b with smallest a.
For this issue, there is a Kadane’s algorithm for maximizing sum: take values greedily and reset count when sum < 0.
Important customization is to always consider countB at least 1 as it must be present in a subarray.

Approach

• we can use Set of only the chars in s

• iterate in ab and ba pairs

• Kotlin API helps save some LOC

Complexity

• Time complexity:
  O(n)

• Space complexity:
  O(n), or O(1) if asSequence used

Code

fun largestVariance(s: String): Int = s.toSet()
.let { ss -> ss.map { a -> ss.filter { it != a }.map { a to it } }.flatten() }
.map { (a, b) ->
    var countA = 0
    var countB = 0
    s.filter { it == a || it == b }
    .map { c ->
        if (c == a) countA++ else countB++
        if (countA < countB) {
            countA = 0
            countB = 0
        }
        countA - maxOf(1, countB)
    }.max() ?: 0
}.max() ?: 0