Smallest Substring With Identical Characters I

HARD

Description

You are given a binary string s of length n and an integer numOps.

You are allowed to perform the following operation on s at most numOps times:

Select any index i (where 0 <= i < n) and flip s[i]. If s[i] == '1', change s[i] to '0' and vice versa.

You need to minimize the length of the longest substring of s such that all the characters in the substring are identical.

Return the minimum length after the operations.

Example 1:

Input: s = "000001", numOps = 1

Output: 2

Explanation:

By changing s[2] to '1', s becomes "001001". The longest substrings with identical characters are s[0..1] and s[3..4].

Example 2:

Input: s = "0000", numOps = 2

Output: 1

Explanation:

By changing s[0] and s[2] to '1', s becomes "1010".

Example 3:

Input: s = "0101", numOps = 0

Output: 1

Constraints:

1 <= n == s.length <= 1000
s consists only of '0' and '1'.
0 <= numOps <= n

Approaches

Checkout 2 different approaches to solve Smallest Substring With Identical Characters I. Click on different approaches to view the approach and algorithm in detail.

Linear Scan on the Answer

This approach involves checking every possible answer for the minimum length, from 1 up to n. For each potential length k, we calculate the minimum number of flips required to ensure no substring of identical characters is longer than k. The first value of k that can be achieved with at most numOps flips is our answer.

Algorithm

Loop through k from 1 to n. k represents the potential maximum length of an identical character substring.
Inside the loop, calculate the total flips needed, requiredFlips, to satisfy the length constraint k.
To calculate requiredFlips:
- Initialize requiredFlips = 0.
- Iterate through the string s to find all contiguous runs of '0's. For each run of length len, if len > k, add len / (k + 1) to requiredFlips.
- Iterate through the string s again to find all contiguous runs of '1's. For each run of length len, if len > k, add len / (k + 1) to requiredFlips.
If requiredFlips <= numOps, it means we can achieve a maximum run length of k. Since we are iterating k from 1 upwards, this is the smallest possible length. Return k.

The core idea is to test each possible length k starting from 1. Since we are looking for the minimum length, the first k that satisfies the condition will be the answer.

To check if a length k is achievable, we need to calculate the minimum number of operations. This involves breaking up any run of identical characters (both '0's and '1's) that is longer than k.

For a run of identical characters of length len > k, we need to introduce flips to break it into segments of length at most k. The most efficient way to do this is to place a flip at every (k+1)-th position. The number of flips required for this single run is floor(len / (k + 1)).

We calculate the total flips needed by summing up the flips required for all runs of '0's longer than k and all runs of '1's longer than k.

We iterate k from 1 to n. For each k, we compute this total required flips. If it's less than or equal to numOps, we've found our minimal length, and we can return k.

class Solution {
    public int smallestSubstring(String s, int numOps) {
        int n = s.length();
        for (int k = 1; k <= n; k++) {
            if (isPossible(s, numOps, k)) {
                return k;
            }
        }
        return n; // Should not be reached given the problem constraints
    }

    private int countFlipsForChar(String s, int k, char targetChar) {
        int flips = 0;
        int currentRun = 0;
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == targetChar) {
                currentRun++;
            } else {
                if (currentRun > k) {
                    flips += currentRun / (k + 1);
                }
                currentRun = 0;
            }
        }
        if (currentRun > k) {
            flips += currentRun / (k + 1);
        }
        return flips;
    }

    private boolean isPossible(String s, int numOps, int k) {
        int requiredFlips = countFlipsForChar(s, k, '0') + countFlipsForChar(s, k, '1');
        return requiredFlips <= numOps;
    }
}

Complexity Analysis

Time Complexity: O(n^2). The outer loop runs up to `n` times. Inside the loop, the `isPossible` check function takes `O(n)` time to scan the string.Space Complexity: O(1). We only use a few variables to keep track of counts and loop indices.

Pros and Cons

Pros:

Simple and straightforward to understand and implement.
Correctly solves the problem for smaller constraints.

Cons:

Inefficient due to the nested loop structure, leading to a quadratic time complexity.
For large n, this approach can be too slow and may result in a 'Time Limit Exceeded' error on some platforms.

Code Solutions

Checking out 3 solutions in different languages for Smallest Substring With Identical Characters I. Click on different languages to view the code.

class Solution {
private
  char[] s;
private
  int numOps;
public
  int minLength(String s, int numOps) {
    this.numOps = numOps;
    this.s = s.toCharArray();
    int l = 1, r = s.length();
    while (l < r) {
      int mid = (l + r) >> 1;
      if (check(mid)) {
        r = mid;
      } else {
        l = mid + 1;
      }
    }
    return l;
  }
private
  boolean check(int m) {
    int cnt = 0;
    if (m == 1) {
      char[] t = {'0', '1'};
      for (int i = 0; i < s.length; ++i) {
        if (s[i] == t[i & 1]) {
          ++cnt;
        }
      }
      cnt = Math.min(cnt, s.length - cnt);
    } else {
      int k = 0;
      for (int i = 0; i < s.length; ++i) {
        ++k;
        if (i == s.length - 1 || s[i] != s[i + 1]) {
          cnt += k / (m + 1);
          k = 0;
        }
      }
    }
    return cnt <= numOps;
  }
}

Video Solution

Watch the video walkthrough for Smallest Substring With Identical Characters I

Algorithms:

Binary Search

Patterns:

Enumeration

Data Structures:

Array