Facebook Pixel

3398. Smallest Substring With Identical Characters I

HardArrayBinary SearchEnumeration
Leetcode Link

Problem Description

You are given a binary string s of length n and an integer numOps. You can 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. The goal is to minimize the length of the longest substring of s such that all the characters in the substring are identical. You need to return the minimum length possible after performing the given operations.

Intuition

The task is to minimize the length of the longest contiguous substring of identical characters by flipping certain characters within the allowed number of operations, numOps. Our aim is to utilize the flips to break up long sequences of identical characters as effectively as possible.

The underlying principle is to consider the length m of the largest contiguous sequence that we can reduce to the minimal required size. To determine if a certain length m can be achieved, we count the number of flips required to ensure no sequence exceeds the length m.

  • Consider lengths sequentially using binary search to find the smallest m that can be achieved.
  • For a given length m, iterate over the string to determine the number of operations required to ensure every contiguous identical section of characters is broken down to m or less.
  • Use a helper function check to see if numOps flips are sufficient for the desired length m.
  • The optimal solution uses binary search (bisect_left) over possible lengths to efficiently find this minimized maximum length.

This approach offers a balanced method leveraging efficient checking of all potential segment modifications while minimizing the larger segments of identical characters.

Learn more about Binary Search patterns.

Solution Approach

The solution employs a binary search approach on the possible lengths of the longest contiguous identical-character sequences that can be formed after making numOps flips. This is a key part of optimizing the problem, as it allows us to efficiently narrow down the smallest possible maximum length.

  1. Binary Search:

    • We perform binary search over the range from 1 to n (the length of the string), treating each value as a potential maximum length of identical-character substrings after the requisite flips.
    • The function bisect_left from the bisect module is used. It finds the minimal length feasible with the given flips.

  2. Helper Function check(m):

    • This function determines if it is feasible to achieve a maximum length of m for contiguous substrings of identical characters using at most numOps flips.
    • It maintains a running count cnt of the flips required:
      • For m = 1, it considers alternating between '0' and '1' to minimize flip cost in two alternative patterns (0101... and 1010...).
      • For m > 1, it uses a counter k to track lengths of contiguous blocks of identical characters and calculates how many complete blocks of size greater than m can be reduced using flips.

  3. Iterating Over Potential Lengths:

    • For each iteration within the range, check(m) is called to see if the current m can indeed be achieved given the flip constraints.
    • The binary search narrows down to the minimum m where check(m) is true, indicating this is the smallest possible maximum length for any substring with identical characters.

This combination of binary search and careful counting through the check function ensures the approach is both efficient and comprehensive in minimizing the longest sequence of identical characters possible with the given constraints.

Ready to land your dream job?

Unlock your dream job with a 2-minute evaluator for a personalized learning plan!

Start Evaluator

Example Walkthrough

Let's illustrate the solution approach with a simple example. Consider the binary string s = "1100101" and numOps = 2. Our goal is to minimize the longest contiguous substring of identical characters using at most 2 flips.

Step-by-Step Process:

  1. Initial Setup:

    • The length of the string n is 7. We will perform a binary search to find the smallest possible length m for the longest sequence of identical characters after performing flips.

  2. Binary Search on Possible Lengths:

    • Start with the range of possible lengths from 1 to 7.
    • Initialize the binary search to determine the smallest possible maximum contiguous length of identical characters.

  3. Checking Length Feasibility (check function):

    • For a given m, check if using numOps = 2 flips allows us to ensure no contiguous section is longer than m.

  4. Testing Example Values:

    • Suppose m = 3:
      • We segment s and evaluate the flips needed to ensure no segment exceeds a length of 3.
      • For s = "1100101", break it into segments: 11, 00, 1, 0, 1.
      • Since no segments exceed length 3, we do not need any flips here, so m = 3 is feasible.

  5. Narrowing Down the Binary Search:

    • As m = 3 is feasible, check smaller lengths.
    • Check for m = 2:
      • Re-evaluate the sequence 11, 00, 1, 0, 1.
      • For segments 11 and 00, flips are required. We can flip 11 to 1X and 00 to 1X (using 1 flip each).
      • With 2 flips available, ensuring all segments are of length 2 or less is feasible.

  6. Finalize the Result:

    • Continue the binary search until we find the smallest m where check(m) is still true.
    • Conclude with the smallest feasible m where the longest contiguous identical segment doesn't exceed this length after flips.

In this example, the final result would be m = 2, as subsequent checks can prove it is the minimum achievable with up to 2 flips.

Solution Implementation

1from bisect import bisect_left
2
3class Solution:
4    def minLength(self, s: str, numOps: int) -> int:
5        def check(m: int) -> bool:
6            # Initialize the counter for operations
7            cnt = 0
8
9            # If m is 1, handle the special case for alternating pattern "01"
10            if m == 1:
11                pattern = "01"
12                # Calculate the minimum operations to make either "010101..." or "101010..."
13                cnt = sum(c == pattern[i % 2] for i, c in enumerate(s))
14                cnt = min(cnt, n - cnt)
15            else:
16                # Otherwise, calculate operations needed for non-alternating sequences
17                k = 0  # Counter for sequence length
18                for i, c in enumerate(s):
19                    k += 1
20                    # If end of sequence or different character, update the counter
21                    if i == len(s) - 1 or c != s[i + 1]:
22                        cnt += k // (m + 1)
23                        k = 0  # Reset sequence length counter
24
25            # Check if the number of operations needed is within the allowed range
26            return cnt <= numOps
27
28        # Length of string s
29        n = len(s)
30      
31        # Use bisect_left to find the smallest m such that check(m) is True
32        return bisect_left(range(n), True, lo=1, key=check)
33
1class Solution {
2    private char[] charArray;
3    private int numOps;
4
5    public int minLength(String s, int numOps) {
6        // Initialize instance variables
7        this.numOps = numOps;
8        this.charArray = s.toCharArray();
9
10        // Binary search initialization
11        int left = 1;
12        int right = s.length();
13
14        // Perform binary search to find the minimum length
15        while (left < right) {
16            int mid = (left + right) >> 1; // Calculate mid-point
17
18            // Check if mid length can be achieved within numOps operations
19            if (check(mid)) {
20                right = mid; // Narrow down the search range
21            } else {
22                left = mid + 1; // Increase the lower bound
23            }
24        }
25
26        // Return the calculated minimum length
27        return left;
28    }
29
30    private boolean check(int m) {
31        int count = 0;
32
33        // Special case when m is 1
34        if (m == 1) {
35            // Two possible alternating patterns: '010101...' and '101010...'
36            char[] pattern = {'0', '1'};
37
38            // Count mismatches for alternating pattern
39            for (int i = 0; i < charArray.length; ++i) {
40                if (charArray[i] == pattern[i & 1]) {
41                    ++count; // Increment count for mismatches
42                }
43            }
44
45            // Calculate the minimum number of changes needed
46            count = Math.min(count, charArray.length - count);
47        } else {
48            // Variable to count consecutive identical characters
49            int k = 0;
50
51            // Iterate the character array to count required operations
52            for (int i = 0; i < charArray.length; ++i) {
53                ++k;
54
55                // Reset k and update count when sequence of same characters ends
56                if (i == charArray.length - 1 || charArray[i] != charArray[i + 1]) {
57                    count += k / (m + 1); // Calculate changes needed
58                    k = 0; // Reset k for new sequence
59                }
60            }
61        }
62
63        // Determine if the count of operations is within the allowed limit
64        return count <= numOps;
65    }
66}
67
1class Solution {
2public:
3    int minLength(string s, int numOps) {
4        int n = s.size();  // Length of the input string
5
6        // Lambda function to check if a particular length `m` can be achieved
7        auto check = [&](int m) {
8            int count = 0;  // To count the number of operations needed
9
10            if (m == 1) {
11                // Special case when m is 1, we check alternating pattern "01"
12                string pattern = "01";
13                for (int i = 0; i < n; ++i) {
14                    // Check if current character matches the alternating pattern
15                    if (s[i] == pattern[i & 1]) {
16                        ++count;
17                    }
18                }
19                // Choose the minimum between reversing `count` or `n - count`
20                count = min(count, n - count);
21            } else {
22                int k = 0;  // Counter for consecutive characters
23                for (int i = 0; i < n; ++i) {
24                    ++k;  // Increment consecutive character counter
25                    // Check if it's the last character or the current and next character differ
26                    if (i == n - 1 || s[i] != s[i + 1]) {
27                        // Calculate number of operations needed
28                        count += k / (m + 1);
29                        k = 0;  // Reset counter for the next sequence
30                    }
31                }
32            }
33
34            // Return true if number of operations is feasible
35            return count <= numOps;
36        };
37
38        int left = 1, right = n;  // Binary search boundaries
39        while (left < right) {
40            int mid = (left + right) >> 1;
41            if (check(mid)) {
42                right = mid;  // Try a smaller length if possible
43            } else {
44                left = mid + 1;  // Move to larger lengths
45            }
46        }
47
48        return left;  // Minimum length found
49    }
50};
51
1function minLength(s: string, numOps: number): number {
2    const n = s.length;
3  
4    // Method to check if it's possible to reduce string length to 'm' using 'numOps' operations
5    const check = (m: number): boolean => {
6        let cnt = 0;
7      
8        if (m === 1) {
9            // Alternate pattern check for 'm' equal to 1
10            const pattern = '01';
11            for (let i = 0; i < n; ++i) {
12                // Count mismatches with the alternating pattern
13                if (s[i] === pattern[i & 1]) {
14                    ++cnt;
15                }
16            }
17            // Find minimum changes needed between matching the pattern or flipping it
18            cnt = Math.min(cnt, n - cnt);
19      
20        } else {
21            // Count complete sequences of length 'm + 1' where repetition occurs
22            let sequenceCount = 0;
23            for (let i = 0; i < n; ++i) {
24                ++sequenceCount;
25                // Check end of sequence or mismatch in sequence
26                if (i === n - 1 || s[i] !== s[i + 1]) {
27                    cnt += Math.floor(sequenceCount / (m + 1));
28                    sequenceCount = 0;
29                }
30            }
31        }
32      
33        // Return true if cnt of operations needed is less than or equal to available operations
34        return cnt <= numOps;
35    };
36  
37    let left = 1;
38    let right = n;
39  
40    // Binary search to find the minimum length possible
41    while (left < right) {
42        const mid = (left + right) >> 1;
43      
44        // If mid length is achievable, try shorter length
45        if (check(mid)) {
46            right = mid;
47        } else {
48            // Otherwise, try longer length
49            left = mid + 1;
50        }
51    }
52  
53    // Return the minimum length found
54    return left;
55}
56

Time and Space Complexity

The time complexity of the minLength function primarily depends on the bisect_left function, which is invoked with a complexity of O(n log n) where n is the length of the string s. Inside the check function, there can be a full traversal of the string s, which takes O(n). Therefore, the overall time complexity of the function can be considered as O(n^2 log n).

The space complexity of the code is O(1) since only a fixed amount of extra space is used for variables, and no additional data structures are used that scale with the input size n.

Learn more about how to find time and space complexity quickly.

Discover Your Strengths and Weaknesses: Take Our 2-Minute Quiz to Tailor Your Study Plan:
Question 1 out of 10

What does the following code do?

1def f(arr1, arr2):
2  i, j = 0, 0
3  new_arr = []
4  while i < len(arr1) and j < len(arr2):
5      if arr1[i] < arr2[j]:
6          new_arr.append(arr1[i])
7          i += 1
8      else:
9          new_arr.append(arr2[j])
10          j += 1
11  new_arr.extend(arr1[i:])
12  new_arr.extend(arr2[j:])
13  return new_arr
14
1public static List<Integer> f(int[] arr1, int[] arr2) {
2  int i = 0, j = 0;
3  List<Integer> newArr = new ArrayList<>();
4
5  while (i < arr1.length && j < arr2.length) {
6      if (arr1[i] < arr2[j]) {
7          newArr.add(arr1[i]);
8          i++;
9      } else {
10          newArr.add(arr2[j]);
11          j++;
12      }
13  }
14
15  while (i < arr1.length) {
16      newArr.add(arr1[i]);
17      i++;
18  }
19
20  while (j < arr2.length) {
21      newArr.add(arr2[j]);
22      j++;
23  }
24
25  return newArr;
26}
27
1function f(arr1, arr2) {
2  let i = 0, j = 0;
3  let newArr = [];
4  
5  while (i < arr1.length && j < arr2.length) {
6      if (arr1[i] < arr2[j]) {
7          newArr.push(arr1[i]);
8          i++;
9      } else {
10          newArr.push(arr2[j]);
11          j++;
12      }
13  }
14  
15  while (i < arr1.length) {
16      newArr.push(arr1[i]);
17      i++;
18  }
19  
20  while (j < arr2.length) {
21      newArr.push(arr2[j]);
22      j++;
23  }
24  
25  return newArr;
26}
27

Recommended Readings

Binary Search Speedrun

Binary Search Speedrun For each of the Speedrun questions you will be given a binary search related problem and a corresponding multiple choice question The multiple choice questions are related to the techniques and template s introduced in the binary search section It's recommended that you have gone through at

Coding Interview Patterns: The Smart, Data-Driven Way to Prepare for Tech Interviews

Coding Interview Patterns Your Personal Dijkstra's Algorithm to Landing Your Dream Job The goal of AlgoMonster is to help you get a job in the shortest amount of time possible in a data driven way We compiled datasets of tech interview problems and broke them down by patterns This way

Recursion Review

Recursion Recursion is one of the most important concepts in computer science Simply speaking recursion is the process of a function calling itself Using a real life analogy imagine a scenario where you invite your friends to lunch https algomonster s3 us east 2 amazonaws com recursion jpg You first

Want a Structured Path to Master System Design Too? Don’t Miss This!

Load More