Problem

Given an integer x (interpreted as a bit string) and a zero-based bit index pos, toggle the bit at position pos without affecting other bits and return the resulting value. Toggling flips the bit: 0 -> 1 and 1 -> 0.

Examples

Example 1

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Input: x = 0b0111, pos = 3
Output: 0b1111
Explanation: toggle the left-most bit (position 3) using a single-bit mask.
Similar Problems

Solution

Method 1 — XOR with a single-bit mask (recommended)

Intuition

XORing a bit with 1 flips it and XORing with 0 leaves it unchanged. Build a mask that has a 1 only at pos and XOR it with x to toggle that bit while leaving others intact.

Approach

  • Create a mask with mask = 1 << pos.
  • Compute result = x ^ mask and return it.
  • Validate pos if needed (e.g., ensure pos >= 0 and within the word width).

Edge cases:

  • If pos is outside the intended bit-width, either normalize (mask pos) or raise/handle as an error depending on caller expectations.

Code

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class Solution {
 public:
  unsigned int toggleBit(unsigned int x, unsigned pos) {
    unsigned int mask = 1u << pos;
    return x ^ mask;
  }
};
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package main

func toggleBit(x uint32, pos uint) uint32 {
    mask := uint32(1) << pos
    return x ^ mask
}
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class Solution {
  public int toggleBit(int x, int pos) {
    int mask = 1 << pos;
    return x ^ mask;
  }
}
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class Solution:
    def toggle_bit(self, x: int, pos: int) -> int:
        mask = 1 << pos
        return x ^ mask

Complexity

  • ⏰ Time complexity: O(1), one bit-shift and one XOR.
  • 🧺 Space complexity: O(1).

Method 2 — Conditional set/clear (explicit)

Intuition

Read the current bit at pos. If it is 1, clear it; if it is 0, set it. This uses mask and bitwise set/clear operations explicitly rather than XOR.

Approach

  • Extract b = (x >> pos) & 1.
  • If b == 1, clear the bit: result = x & ~(1 << pos).
  • Else set the bit: result = x | (1 << pos).

Code

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class Solution {
 public:
  unsigned int toggleBitExplicit(unsigned int x, unsigned pos) {
    unsigned int b = (x >> pos) & 1u;
    if (b == 1u) return x & ~(1u << pos);
    return x | (1u << pos);
  }
};
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package main

func toggleBitExplicit(x uint32, pos uint) uint32 {
    b := (x >> pos) & 1
    if b == 1 {
        return x & ^(uint32(1) << pos)
    }
    return x | (uint32(1) << pos)
}
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class Solution {
  public int toggleBitExplicit(int x, int pos) {
    int b = (x >> pos) & 1;
    if (b == 1) return x & ~(1 << pos);
    return x | (1 << pos);
  }
}
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class Solution:
    def toggle_bit_explicit(self, x: int, pos: int) -> int:
        b = (x >> pos) & 1
        if b == 1:
            return x & ~(1 << pos)
        return x | (1 << pos)

Complexity

  • ⏰ Time complexity: O(1).
  • 🧺 Space complexity: O(1).

Notes

  • This is a basic bit-manipulation primitive. There is no exact LeetCode problem with this precise title — it’s a commonly used operation in many bitwise tasks.
  • Original reference kept in source_links.