LeetCode #3777 — HARD

Minimum Deletions to Make Alternating Substring

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given a string s of length n consisting only of the characters 'A' and 'B'.

You are also given a 2D integer array queries of length q, where each queries[i] is one of the following:

[1, j]: Flip the character at index j of s i.e. 'A' changes to 'B' (and vice versa). This operation mutates s and affects subsequent queries.
[2, l, r]: Compute the minimum number of character deletions required to make the substring s[l..r] alternating. This operation does not modify s; the length of s remains n.

A substring is alternating if no two adjacent characters are equal. A substring of length 1 is always alternating.

Return an integer array answer, where answer[i] is the result of the i^th query of type [2, l, r].

Example 1:

Input: s = "ABA", queries = [[2,1,2],[1,1],[2,0,2]]

Output: [0,2]

Explanation:

`i`	`queries[i]`	`j`	`l`	`r`	`s` before query	`s[l..r]`	Result	Answer
0	[2, 1, 2]	-	1	2	`"ABA"`	`"BA"`	Already alternating	0
1	[1, 1]	1	-	-	`"ABA"`	-	Flip `s[1]` from `'B'` to `'A'`	-
2	[2, 0, 2]	-	0	2	`"AAA"`	`"AAA"`	Delete any two `'A'`s to get `"A"`	2

Thus, the answer is [0, 2].

Example 2:

Input: s = "ABB", queries = [[2,0,2],[1,2],[2,0,2]]

Output: [1,0]

Explanation:

`i`	`queries[i]`	`j`	`l`	`r`	`s` before query	`s[l..r]`	Result	Answer
0	[2, 0, 2]	-	0	2	`"ABB"`	`"ABB"`	Delete one `'B'` to get `"AB"`	1
1	[1, 2]	2	-	-	`"ABB"`	-	Flip `s[2]` from `'B'` to `'A'`	-
2	[2, 0, 2]	-	0	2	`"ABA"`	`"ABA"`	Already alternating	0

Thus, the answer is [1, 0].

Example 3:

Input: s = "BABA", queries = [[2,0,3],[1,1],[2,1,3]]

Output: [0,1]

Explanation:

`i`	`queries[i]`	`j`	`l`	`r`	`s` before query	`s[l..r]`	Result	Answer
0	[2, 0, 3]	-	0	3	`"BABA"`	`"BABA"`	Already alternating	0
1	[1, 1]	1	-	-	`"BABA"`	-	Flip `s[1]` from `'A'` to `'B'`	-
2	[2, 1, 3]	-	1	3	`"BBBA"`	`"BBA"`	Delete one `'B'` to get `"BA"`	1

Thus, the answer is [0, 1].

Constraints:

1 <= n == s.length <= 10⁵
s[i] is either 'A' or 'B'.
1 <= q == queries.length <= 10⁵
queries[i].length == 2 or 3
- queries[i] == [1, j] or,
- queries[i] == [2, l, r]
- 0 <= j <= n - 1
- 0 <= l <= r <= n - 1

Step 01

Brute Force Baseline

Problem summary: You are given a string s of length n consisting only of the characters 'A' and 'B'. You are also given a 2D integer array queries of length q, where each queries[i] is one of the following: [1, j]: Flip the character at index j of s i.e. 'A' changes to 'B' (and vice versa). This operation mutates s and affects subsequent queries. [2, l, r]: Compute the minimum number of character deletions required to make the substring s[l..r] alternating. This operation does not modify s; the length of s remains n. A substring is alternating if no two adjacent characters are equal. A substring of length 1 is always alternating. Return an integer array answer, where answer[i] is the result of the ith query of type [2, l, r].

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Segment Tree

Example 1

"ABA"
[[2,1,2],[1,1],[2,0,2]]

Example 2

"ABB"
[[2,0,2],[1,2],[2,0,2]]

Example 3

"BABA"
[[2,0,3],[1,1],[2,1,3]]

Step 02

Core Insight

What unlocks the optimal approach

Use a Fenwick tree (BIT) over an auxiliary array <code>eq</code>.
Define <code>eq[i] = 1</code> if <code>i >= 1</code> and <code>s[i] == s[i - 1]</code>, otherwise <code>eq[i] = 0</code>.
For a type-2 query <code>[2, l, r]</code> the answer is <code>sum(eq[l+1..r])</code> (count of equal adjacent pairs in the substring).
For a flip <code>[1, j]</code>, recompute and update <code>eq[j]</code> and <code>eq[j + 1]</code>; each flip changes at most two <code>eq</code> values.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3777: Minimum Deletions to Make Alternating Substring
class BinaryIndexedTree {
    int n;
    int[] c;

    BinaryIndexedTree(int n) {
        this.n = n;
        this.c = new int[n + 1];
    }

    void update(int x, int delta) {
        while (x <= n) {
            c[x] += delta;
            x += x & -x;
        }
    }

    int query(int x) {
        int s = 0;
        while (x > 0) {
            s += c[x];
            x -= x & -x;
        }
        return s;
    }
}

class Solution {
    public int[] minDeletions(String s, int[][] queries) {
        int n = s.length();
        int[] nums = new int[n];
        BinaryIndexedTree bit = new BinaryIndexedTree(n);

        for (int i = 1; i < n; i++) {
            nums[i] = (s.charAt(i) == s.charAt(i - 1)) ? 1 : 0;
            if (nums[i] == 1) {
                bit.update(i + 1, 1);
            }
        }

        int cnt = 0;
        for (int[] q : queries) {
            if (q[0] == 2) {
                cnt++;
            }
        }

        int[] ans = new int[cnt];
        int idx = 0;

        for (int[] q : queries) {
            if (q[0] == 1) {
                int j = q[1];

                int delta = (nums[j] ^ 1) - nums[j];
                nums[j] ^= 1;
                bit.update(j + 1, delta);

                if (j + 1 < n) {
                    delta = (nums[j + 1] ^ 1) - nums[j + 1];
                    nums[j + 1] ^= 1;
                    bit.update(j + 2, delta);
                }
            } else {
                int l = q[1];
                int r = q[2];
                ans[idx++] = bit.query(r + 1) - bit.query(l + 1);
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3777: Minimum Deletions to Make Alternating Substring
type binaryIndexedTree struct {
	n int
	c []int
}

func newBinaryIndexedTree(n int) *binaryIndexedTree {
	return &binaryIndexedTree{
		n: n,
		c: make([]int, n+1),
	}
}

func (bit *binaryIndexedTree) update(x, delta int) {
	for x <= bit.n {
		bit.c[x] += delta
		x += x & -x
	}
}

func (bit *binaryIndexedTree) query(x int) int {
	s := 0
	for x > 0 {
		s += bit.c[x]
		x -= x & -x
	}
	return s
}

func minDeletions(s string, queries [][]int) []int {
	n := len(s)
	nums := make([]int, n)
	bit := newBinaryIndexedTree(n)

	for i := 1; i < n; i++ {
		if s[i] == s[i-1] {
			nums[i] = 1
			bit.update(i+1, 1)
		}
	}

	ans := make([]int, 0)

	for _, q := range queries {
		if q[0] == 1 {
			j := q[1]

			delta := (nums[j] ^ 1 - nums[j])
			nums[j] ^= 1
			bit.update(j+1, delta)

			if j+1 < n {
				delta = (nums[j+1] ^ 1 - nums[j+1])
				nums[j+1] ^= 1
				bit.update(j+2, delta)
			}
		} else {
			l, r := q[1], q[2]
			ans = append(ans, bit.query(r+1)-bit.query(l+1))
		}
	}

	return ans
}

# Accepted solution for LeetCode #3777: Minimum Deletions to Make Alternating Substring
class BinaryIndexedTree:
    __slots__ = "n", "c"

    def __init__(self, n: int):
        self.n = n
        self.c = [0] * (n + 1)

    def update(self, x: int, delta: int) -> None:
        while x <= self.n:
            self.c[x] += delta
            x += x & -x

    def query(self, x: int) -> int:
        s = 0
        while x:
            s += self.c[x]
            x -= x & -x
        return s


class Solution:
    def minDeletions(self, s: str, queries: List[List[int]]) -> List[int]:
        n = len(s)
        nums = [0] * n
        bit = BinaryIndexedTree(n)
        for i in range(1, n):
            nums[i] = int(s[i] == s[i - 1])
            if nums[i]:
                bit.update(i + 1, 1)
        ans = []
        for q in queries:
            if q[0] == 1:
                j = q[1]
                delta = (nums[j] ^ 1) - nums[j]
                nums[j] ^= 1
                bit.update(j + 1, delta)
                if j + 1 < n:
                    delta = (nums[j + 1] ^ 1) - nums[j + 1]
                    nums[j + 1] ^= 1
                    bit.update(j + 2, delta)
            else:
                _, l, r = q
                ans.append(bit.query(r + 1) - bit.query(l + 1))
        return ans

// Accepted solution for LeetCode #3777: Minimum Deletions to Make Alternating Substring
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3777: Minimum Deletions to Make Alternating Substring
// class BinaryIndexedTree {
//     int n;
//     int[] c;
// 
//     BinaryIndexedTree(int n) {
//         this.n = n;
//         this.c = new int[n + 1];
//     }
// 
//     void update(int x, int delta) {
//         while (x <= n) {
//             c[x] += delta;
//             x += x & -x;
//         }
//     }
// 
//     int query(int x) {
//         int s = 0;
//         while (x > 0) {
//             s += c[x];
//             x -= x & -x;
//         }
//         return s;
//     }
// }
// 
// class Solution {
//     public int[] minDeletions(String s, int[][] queries) {
//         int n = s.length();
//         int[] nums = new int[n];
//         BinaryIndexedTree bit = new BinaryIndexedTree(n);
// 
//         for (int i = 1; i < n; i++) {
//             nums[i] = (s.charAt(i) == s.charAt(i - 1)) ? 1 : 0;
//             if (nums[i] == 1) {
//                 bit.update(i + 1, 1);
//             }
//         }
// 
//         int cnt = 0;
//         for (int[] q : queries) {
//             if (q[0] == 2) {
//                 cnt++;
//             }
//         }
// 
//         int[] ans = new int[cnt];
//         int idx = 0;
// 
//         for (int[] q : queries) {
//             if (q[0] == 1) {
//                 int j = q[1];
// 
//                 int delta = (nums[j] ^ 1) - nums[j];
//                 nums[j] ^= 1;
//                 bit.update(j + 1, delta);
// 
//                 if (j + 1 < n) {
//                     delta = (nums[j + 1] ^ 1) - nums[j + 1];
//                     nums[j + 1] ^= 1;
//                     bit.update(j + 2, delta);
//                 }
//             } else {
//                 int l = q[1];
//                 int r = q[2];
//                 ans[idx++] = bit.query(r + 1) - bit.query(l + 1);
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3777: Minimum Deletions to Make Alternating Substring
class BinaryIndexedTree {
    n: number;
    c: number[];

    constructor(n: number) {
        this.n = n;
        this.c = Array(n + 1).fill(0);
    }

    update(x: number, delta: number): void {
        while (x <= this.n) {
            this.c[x] += delta;
            x += x & -x;
        }
    }

    query(x: number): number {
        let s = 0;
        while (x > 0) {
            s += this.c[x];
            x -= x & -x;
        }
        return s;
    }
}

function minDeletions(s: string, queries: number[][]): number[] {
    const n = s.length;
    const nums: number[] = Array(n).fill(0);
    const bit = new BinaryIndexedTree(n);

    for (let i = 1; i < n; i++) {
        if (s[i] === s[i - 1]) {
            nums[i] = 1;
            bit.update(i + 1, 1);
        }
    }

    const ans: number[] = [];

    for (const q of queries) {
        if (q[0] === 1) {
            const j = q[1];

            let delta = (nums[j] ^ 1) - nums[j];
            nums[j] ^= 1;
            bit.update(j + 1, delta);

            if (j + 1 < n) {
                delta = (nums[j + 1] ^ 1) - nums[j + 1];
                nums[j + 1] ^= 1;
                bit.update(j + 2, delta);
            }
        } else {
            const l = q[1],
                r = q[2];
            ans.push(bit.query(r + 1) - bit.query(l + 1));
        }
    }

    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O((n + q)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n × q) time

O(1) space

For each of q queries, scan the entire range to compute the aggregate (sum, min, max). Each range scan takes up to O(n) for a full-array query. With q queries: O(n × q) total. Point updates are O(1) but queries dominate.

SEGMENT TREE

O(n + q log n) time

O(n) space

Building the tree is O(n). Each query or update traverses O(log n) nodes (tree height). For q queries: O(n + q log n) total. Space is O(4n) ≈ O(n) for the tree array. Lazy propagation adds O(1) per node for deferred updates.

Shortcut: Build O(n), query/update O(log n) each. When you need both range queries AND point updates.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.