1963. Minimum Number of Swaps to Make The String Balanced - Explanation

Prerequisites

Before attempting this problem, you should be comfortable with:

Stack Data Structure - Used to track unmatched opening brackets during traversal
Bracket Matching - Understanding how to pair opening and closing brackets in sequence
Greedy Algorithms - The optimal solutions use greedy reasoning to count minimum swaps needed

1. Stack

Intuition

A balanced string has every ] matched with a preceding [. We use a stack to track unmatched opening brackets. When we see [, we push it. When we see ] and the stack is not empty, we pop (the bracket is matched). If the stack is empty when we see ], that closing bracket is unmatched.

After processing, the stack contains only unmatched [ brackets. Since the string has equal counts of [ and ], the number of unmatched [ equals the number of unmatched ]. Each swap fixes two unmatched pairs, so we need (unmatched + 1) / 2 swaps.

Algorithm

Initialize an empty stack.
Iterate through the string:
- If the character is [, push it onto the stack.
- If the character is ] and the stack is not empty, pop the stack.
The remaining stack size represents unmatched [ brackets.
Return (stack_size + 1) / 2.

class Solution:
    def minSwaps(self, s: str) -> int:
        stack = []
        for c in s:
            if c == '[':
                stack.append(c)
            elif stack:
                stack.pop()
        return (len(stack) + 1) // 2

public class Solution {
    public int minSwaps(String s) {
        Stack<Character> stack = new Stack<>();
        for (char c : s.toCharArray()) {
            if (c == '[') {
                stack.push(c);
            } else if (!stack.isEmpty()) {
                stack.pop();
            }
        }
        return (stack.size() + 1) / 2;
    }
}

class Solution {
public:
    int minSwaps(string s) {
        vector<char> stack;
        for (char c : s) {
            if (c == '[') {
                stack.push_back(c);
            } else if (!stack.empty()) {
                stack.pop_back();
            }
        }
        return (stack.size() + 1) / 2;
    }
};

class Solution {
    /**
     * @param {string} s
     * @return {number}
     */
    minSwaps(s) {
        let stack = [];
        for (const c of s) {
            if (c === '[') {
                stack.push(c);
            } else if (stack.length > 0) {
                stack.pop();
            }
        }
        return Math.floor((stack.length + 1) / 2);
    }
}

public class Solution {
    public int MinSwaps(string s) {
        Stack<char> stack = new Stack<char>();
        foreach (char c in s) {
            if (c == '[') {
                stack.Push(c);
            } else if (stack.Count > 0) {
                stack.Pop();
            }
        }
        return (stack.Count + 1) / 2;
    }
}

func minSwaps(s string) int {
    stack := 0
    for _, c := range s {
        if c == '[' {
            stack++
        } else if stack > 0 {
            stack--
        }
    }
    return (stack + 1) / 2
}

class Solution {
    fun minSwaps(s: String): Int {
        val stack = ArrayDeque<Char>()
        for (c in s) {
            if (c == '[') {
                stack.addLast(c)
            } else if (stack.isNotEmpty()) {
                stack.removeLast()
            }
        }
        return (stack.size + 1) / 2
    }
}

class Solution {
    func minSwaps(_ s: String) -> Int {
        var stack = [Character]()
        for c in s {
            if c == "[" {
                stack.append(c)
            } else if !stack.isEmpty {
                stack.removeLast()
            }
        }
        return (stack.count + 1) / 2
    }
}

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(n)$

2. Greedy - I

Intuition

Instead of tracking opening brackets, we can track the imbalance directly. We maintain a close counter that increases for ] and decreases for [. The max value this counter reaches tells us the worst-case imbalance, meaning the maximum number of unmatched closing brackets at any point.

Since each swap can fix at most 2 unmatched brackets, the number of swaps needed is (max_imbalance + 1) / 2.

Algorithm

Initialize close = 0 and maxClose = 0.
Iterate through the string:
- If the character is [, decrement close.
- If the character is ], increment close.
- Update maxClose = max(maxClose, close).
Return (maxClose + 1) / 2.

class Solution:
    def minSwaps(self, s: str) -> int:
        close = maxClose = 0

        for c in s:
            if c == '[':
                close -= 1
            else:
                close += 1
            maxClose = max(maxClose, close)

        return (maxClose + 1) // 2

public class Solution {
    public int minSwaps(String s) {
        int close = 0, maxClose = 0;
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == '[') close--;
            else close++;
            maxClose = Math.max(maxClose, close);
        }
        return (maxClose + 1) / 2;
    }
}

class Solution {
public:
    int minSwaps(string s) {
        int close = 0, maxClose = 0;
        for (auto& c : s) {
            if (c == '[') close--;
            else close++;
            maxClose = max(maxClose, close);
        }
        return (maxClose + 1) / 2;
    }
};

class Solution {
    /**
     * @param {string} s
     * @return {number}
     */
    minSwaps(s) {
        let close = 0,
            maxClose = 0;
        for (let i = 0; i < s.length; i++) {
            if (s.charAt(i) == '[') close--;
            else close++;
            maxClose = Math.max(maxClose, close);
        }
        return Math.floor((maxClose + 1) / 2);
    }
}

public class Solution {
    public int MinSwaps(string s) {
        int close = 0, maxClose = 0;
        foreach (char c in s) {
            if (c == '[') close--;
            else close++;
            maxClose = Math.Max(maxClose, close);
        }
        return (maxClose + 1) / 2;
    }
}

func minSwaps(s string) int {
    close, maxClose := 0, 0
    for _, c := range s {
        if c == '[' {
            close--
        } else {
            close++
        }
        if close > maxClose {
            maxClose = close
        }
    }
    return (maxClose + 1) / 2
}

class Solution {
    fun minSwaps(s: String): Int {
        var close = 0
        var maxClose = 0
        for (c in s) {
            if (c == '[') close--
            else close++
            maxClose = maxOf(maxClose, close)
        }
        return (maxClose + 1) / 2
    }
}

class Solution {
    func minSwaps(_ s: String) -> Int {
        var close = 0
        var maxClose = 0
        for c in s {
            if c == "[" {
                close -= 1
            } else {
                close += 1
            }
            maxClose = max(maxClose, close)
        }
        return (maxClose + 1) / 2
    }
}

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(1)$

3. Greedy - II

Intuition

This approach directly simulates the stack without actually using a stack data structure. We use a counter stackSize that increments for [ and decrements for ] only if there is something to match (stackSize > 0). The final counter value represents unmatched opening brackets.

This is equivalent to the stack approach but uses O(1) space since we only track the count, not the actual characters.

Algorithm

Initialize stackSize = 0.
Iterate through the string:
- If the character is [, increment stackSize.
- If the character is ] and stackSize > 0, decrement stackSize.
Return (stackSize + 1) / 2.

class Solution:
    def minSwaps(self, s: str) -> int:
        stackSize = 0
        for c in s:
            if c == '[':
                stackSize += 1
            elif stackSize > 0:
                stackSize -= 1
        return (stackSize + 1) // 2

public class Solution {
    public int minSwaps(String s) {
        int stackSize = 0;
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == '[') stackSize++;
            else if (stackSize > 0) stackSize--;
        }
        return (stackSize + 1) / 2;
    }
}

class Solution {
public:
    int minSwaps(string s) {
        int stackSize = 0;
        for (auto& c : s) {
            if (c == '[') stackSize++;
            else if (stackSize > 0) stackSize--;
        }
        return (stackSize + 1) / 2;
    }
};

class Solution {
    /**
     * @param {string} s
     * @return {number}
     */
    minSwaps(s) {
        let stackSize = 0;
        for (let i = 0; i < s.length; i++) {
            if (s.charAt(i) == '[') stackSize++;
            else if (stackSize > 0) stackSize--;
        }
        return Math.floor((stackSize + 1) / 2);
    }
}

public class Solution {
    public int MinSwaps(string s) {
        int stackSize = 0;
        foreach (char c in s) {
            if (c == '[') stackSize++;
            else if (stackSize > 0) stackSize--;
        }
        return (stackSize + 1) / 2;
    }
}

func minSwaps(s string) int {
    stackSize := 0
    for _, c := range s {
        if c == '[' {
            stackSize++
        } else if stackSize > 0 {
            stackSize--
        }
    }
    return (stackSize + 1) / 2
}

class Solution {
    fun minSwaps(s: String): Int {
        var stackSize = 0
        for (c in s) {
            if (c == '[') stackSize++
            else if (stackSize > 0) stackSize--
        }
        return (stackSize + 1) / 2
    }
}

class Solution {
    func minSwaps(_ s: String) -> Int {
        var stackSize = 0
        for c in s {
            if c == "[" {
                stackSize += 1
            } else if stackSize > 0 {
                stackSize -= 1
            }
        }
        return (stackSize + 1) / 2
    }
}

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(1)$

Common Pitfalls

Counting All Brackets Instead of Unmatched

The answer depends on unmatched brackets, not total brackets. Counting all [ or ] characters without first matching valid pairs will give an incorrect count. Only brackets that remain after matching contribute to the swap count.

Forgetting That One Swap Fixes Two Pairs

Each swap can fix two unmatched bracket pairs simultaneously. Returning the unmatched count directly instead of dividing by 2 (with ceiling) will double the actual number of swaps needed.

Popping From Empty Stack

When simulating with a stack, attempting to pop when encountering ] without checking if the stack is empty will cause runtime errors. Only pop if there is a matching [ available to pair with the current ].