Find maximum of two numbers without using comparison operators or if-else

Problem

Write a method which finds the maximum of two numbers You should not use if-else or any other comparison operator

Examples

Example 1

Input: a = 5, b = 10
Output: 10

Solution

Method 1 - Arithmetic midpoint (overflow-aware)

Intuition

The average-based formula

ans = (a + b) / 2 + |a - b| / 2

returns the larger of a and b. It follows from decomposing max(a,b) as midpoint plus half the absolute difference. This is simple and branch-free, but naive use can overflow when a + b or |a - b| exceed the integer range.

Approach

• Use 64-bit intermediate arithmetic to avoid overflow for 32-bit inputs.
• Compute sum = (long long)a + (long long)b, diff = llabs((long long)a - (long long)b).
• ans = sum/2 + diff/2.
• For min, use sum/2 - diff/2.
• Edge cases: works for negative numbers, equal numbers, and handles typical 32-bit ranges when using 64-bit intermediates; still document limits for extreme inputs beyond 64-bit.

Code

C++

class Solution {
 public:
  int getMax(int a, int b) {
    long long s = static_cast<long long>(a) + static_cast<long long>(b);
    long long d = llabs(static_cast<long long>(a) - static_cast<long long>(b));
    int ans = static_cast<int>(s / 2 + d / 2);
    return ans;
  }

  int getMin(int a, int b) {
    long long s = static_cast<long long>(a) + static_cast<long long>(b);
    long long d = llabs(static_cast<long long>(a) - static_cast<long long>(b));
    int ans = static_cast<int>(s / 2 - d / 2);
    return ans;
  }
};

Go

package main

func getMax(a, b int) int {
    var s int64 = int64(a) + int64(b)
    var d int64
    if a >= b { d = int64(a - b) } else { d = int64(b - a) }
    ans := int(s/2 + d/2)
    return ans
}

func getMin(a, b int) int {
    var s int64 = int64(a) + int64(b)
    var d int64
    if a >= b { d = int64(a - b) } else { d = int64(b - a) }
    ans := int(s/2 - d/2)
    return ans
}

Java

class Solution {
  public int getMax(int a, int b) {
    long s = (long) a + (long) b;
    long d = Math.abs((long) a - (long) b);
    int ans = (int) (s / 2 + d / 2);
    return ans;
  }

  public int getMin(int a, int b) {
    long s = (long) a + (long) b;
    long d = Math.abs((long) a - (long) b);
    int ans = (int) (s / 2 - d / 2);
    return ans;
  }
}

Kotlin

class Solution {
  fun getMax(a: Int, b: Int): Int {
    val s = a.toLong() + b.toLong()
    val d = kotlin.math.abs(a.toLong() - b.toLong())
    val ans = (s / 2 + d / 2).toInt()
    return ans
  }

  fun getMin(a: Int, b: Int): Int {
    val s = a.toLong() + b.toLong()
    val d = kotlin.math.abs(a.toLong() - b.toLong())
    val ans = (s / 2 - d / 2).toInt()
    return ans
  }
}

Python

class Solution:
    def getMax(self, a: int, b: int) -> int:
        s: int = int(a) + int(b)
        d: int = abs(int(a) - int(b))
        ans: int = int(s // 2 + d // 2)
        return ans

    def getMin(self, a: int, b: int) -> int:
        s: int = int(a) + int(b)
        d: int = abs(int(a) - int(b))
        ans: int = int(s // 2 - d // 2)
        return ans

Complexity

• ⏰ Time complexity: O(1), constant-time arithmetic operations for addition, subtraction and division.
• 🧺 Space complexity: O(1), constant extra space for temporaries.

Method 2 - Sign-bit extraction (MSB technique)

Intuition

Compute c = a - b. The sign bit of c indicates whether a < b (sign bit 1) or a >= b (sign bit 0) in two's-complement representation. Use that bit as k (0 or 1) and return a - k * c which equals max(a,b). This is branchless and works on fixed-width two's-complement integers.

Approach

• Compute c = a - b using an integer type that has the same width as inputs.
• Extract the sign bit into k as an unsigned 0/1 (careful with right-shift semantics and signedness).
• ans_max = a - k * c.
• ans_min = b + k * c (equivalently a - (1-k) * c).
• Edge cases: behavior depends on two's-complement representation and defined shift behavior — use unsigned shifts or cast to unsigned to read the high bit portably. Beware of overflow when a - b is outside the signed range for the chosen type.

Code

C++

class Solution {
 public:
  int getMax(int a, int b) {
    int c = a - b;
    unsigned int uc = static_cast<unsigned int>(c);
    int k = static_cast<int>((uc >> 31) & 0x1u);
    int ans = a - k * c;
    return ans;
  }

  int getMin(int a, int b) {
    int c = a - b;
    unsigned int uc = static_cast<unsigned int>(c);
    int k = static_cast<int>((uc >> 31) & 0x1u);
    int ans = b + k * c;
    return ans;
  }
};

Go

package main

func getMax(a, b int) int {
    c := a - b
    // convert to unsigned to perform logical right shift on sign bit
    uc := uint32(int32(c))
    k := int((uc >> 31) & 0x1)
    ans := a - k*c
    return ans
}

func getMin(a, b int) int {
    c := a - b
    uc := uint32(int32(c))
    k := int((uc >> 31) & 0x1)
    ans := b + k*c
    return ans
}

Java

class Solution {
  public int getMax(int a, int b) {
    int c = a - b;
    int k = (c >> 31) & 0x1;
    int ans = a - k * c;
    return ans;
  }

  public int getMin(int a, int b) {
    int c = a - b;
    int k = (c >> 31) & 0x1;
    int ans = b + k * c;
    return ans;
  }
}

Kotlin

class Solution {
  fun getMax(a: Int, b: Int): Int {
    val c = a - b
    val k = (c shr 31) and 0x1
    val ans = a - k * c
    return ans
  }

  fun getMin(a: Int, b: Int): Int {
    val c = a - b
    val k = (c shr 31) and 0x1
    val ans = b + k * c
    return ans
  }
}

Python

class Solution:
    def getMax(self, a: int, b: int) -> int:
        c: int = int(a) - int(b)
        # mask to 32 bits then shift to obtain sign bit as 0/1
        uc: int = c & 0xFFFFFFFF
        k: int = (uc >> 31) & 0x1
        ans: int = int(a) - k * c
        return ans

    def getMin(self, a: int, b: int) -> int:
        c: int = int(a) - int(b)
        uc: int = c & 0xFFFFFFFF
        k: int = (uc >> 31) & 0x1
        ans: int = int(b) + k * c
        return ans

Complexity

• ⏰ Time complexity: O(1), constant-time arithmetic and bit operations.
• 🧺 Space complexity: O(1), constant extra space for a few temporaries.