# Substring With Largest Variance

Published: Sep 14, 2022

Hard Dynamic Programming Array

## Introduction

The basic idea comes up easily for this problem. Create substring and count characters. A difference of maximum and minimum counts are the answer of the substring. However, such brute force solution takes too long time to complete. The key here is how to accelerate the process.

## Problem Description

The variance of a string is defined as the largest difference between the number of occurrences of any 2 characters present in the string. Note the two characters may or may not be the same.

Given a string `s` consisting of lowercase English letters only, return the largest variance possible among all substrings of `s`.

A substring is a contiguous sequence of characters within a string.

Constraints:

• `1 <= s.length <= 10**4`
• `s` consists of lowercase English letters.

https://leetcode.com/problems/substring-with-largest-variance/

## Examples

``````Example 1:
Input: s = "aababbb"
Output: 3
Explanation: "babbb" has a variance 3.
``````
``````Example 2:
Input: s = "abcde"
Output: 0
``````

## Analysis

The solution starts from choosing two letters. Python’s permutation function takes too long for this problem. So, a string of no-duplication is created first. The loop goes over every two letter combinations. Then, using an auxiliary array, calculate local max. This leads to the ground max.

## Solution

``````class SubstringWithLargestVariance:
def largestVariance(self, s: str) -> int:
def maxSubArray(nums: List[int]):
max_v = float('-inf')
runningSum = 0
seen = False
for x in nums:
if x < 0:
seen = True
runningSum += x
if seen:
max_v = max(max_v, runningSum)
else:
max_v = max(max_v, runningSum - 1)
if runningSum < 0:
runningSum = 0
seen = False
return max_v

f = set()
a = ''
for c in s:
if c not in f:
a += c

result = 0
for i in range(len(a) - 1):
for j in range(i + 1,len(a)):
x, y = a[i], a[j]
arr = []
for c in s:
if c != x and c != y:
continue
elif c == x:
arr.append(1)
else:
arr.append(-1)
result = max(result, maxSubArray(arr), maxSubArray([-v for v in arr]))
return result
``````

## Complexities

• Time: `O(n^2)`
• Space: `O(n)`