{"@context":"https://schema.org","@type":"TechArticle","headline":"Sum of Two Integers Algorithm - Visualization & Solution","description":"Given two integers a and b, return the sum of the two integers without using the operators + and -.\n\nExample 1:\nInput: a = 1, b = 2\nOutput: 3\n\nExample 2:\nInput: a = 2, b = 3\nOutput: 5\n\nNotes: Use bitwise operations (XOR and AND with shifts) to simulate addition without using + or -.","image":"https://rulcode.com/og-image.png","inLanguage":"en","datePublished":"2025-12-01T16:02:46.107488+00:00","dateModified":"2026-05-30T13:21:57.287057+00:00","category":" Bit Manipulation, Math","author":{"@type":"Organization","name":"Rulcode.com Team","url":"https://rulcode.com"},"publisher":{"@type":"Organization","name":"Rulcode.com","logo":{"@type":"ImageObject","url":"https://rulcode.com/android-chrome-512x512.png"}},"mainEntityOfPage":{"@type":"WebPage","@id":"https://rulcode.com/problem/sum-of-two-integers"},"about":{"@type":"Thing","name":"Sum of Two Integers","description":"Given two integers a and b, return the sum of the two integers without using the operators + and -.\n\nExample 1:\nInput: a = 1, b = 2\nOutput: 3\n\nExample 2:\nInput: a = 2, b = 3\nOutput: 5\n\nNotes: Use bitwise operations (XOR and AND with shifts) to simulate addition without using + or -."},"proficiencyLevel":"intermediate","hasPart":[{"@context":"https://schema.org","@type":"SoftwareSourceCode","programmingLanguage":"python","codeSampleType":"optimize","text":"def getSum(a: int, b: int) -> int:\n # Loop until there is no carry\n # Python integers are not fixed-width (32-bit)\n MASK = 0xFFFFFFFF # 32 bits - Ensures we work within 32-bit range, simulating integer overflow\n MAX = 0x7FFFFFFF # max positive int - Used to convert to signed 32-bit integer\n\n while b != 0:\n # Carry is produced where both bits are 1\n carry = (a & b) & MASK # Calculate carry by performing bitwise AND and applying the mask\n\n # XOR gives sum without carry\n a = (a ^ b) & MASK # Calculate sum (without carry) using bitwise XOR and applying the mask\n\n # Shift carry left by 1 to add in next position\n b = (carry << 1) & MASK # Shift the carry to the left by 1 bit and apply the mask\n\n # Final result stored in 'a'\n\n # Convert to signed integer\n return a if a <= MAX else ~(a ^ MASK) # Convert to a signed 32-bit integer if necessary. If 'a' is greater than MAX, it's a negative number in 2's complement. We invert and apply the mask.\n","description":"

\n\n This approach adds two integers without using the + or\n - operators by simulating how addition works internally at the\n binary level. Instead of arithmetic operations, it relies on bitwise logic.\n

\n\n

\n Binary addition can be split into two parts: calculating the sum without\n carry and calculating the carry itself. These two parts are combined\n repeatedly until no carry remains.\n

\n\n

\n By looping until the carry becomes zero, this method guarantees that all\n bits are added correctly, producing the final sum efficiently.\n

\n\n

\n Intuition (Simple Language):
\n Think about how you add numbers on paper.\n

\n\n

\n You write the number and carry extra value to the next column.\n Binary addition works the same way.\n

\n\n

\n XOR (^) gives the sum without carry, and\n AND (&) tells us where the carry is created.\n

\n\n

\n The carry belongs to the next higher bit, so we shift it left.\n We keep repeating this until there is no carry left.\n

\n\n

\n When no carry remains, the value left is the final answer.\n

\n\n

\n Steps to Solve:\n

\n\n

Start with two integers a and b.
Calculate the carry using a & b.
Calculate the sum without carry using a ^ b.
Shift the carry left by one bit using << 1.
Assign the XOR result back to a.
Assign the shifted carry back to b.
Repeat until b becomes zero.
Return a as the final sum.

\n\n

\n Time Complexity: O(1)
\n Space Complexity: O(1)
\n
\n\n \n Learn Complexity\n \n
\n
\n

\n"},{"@context":"https://schema.org","@type":"SoftwareSourceCode","programmingLanguage":"python","codeSampleType":"starter","text":"def getSum(a: int, b: int) -> int:\n # TODO: Implement\n return 0\n","description":"Sum of Two Integers implementation in python"},{"@context":"https://schema.org","@type":"SoftwareSourceCode","programmingLanguage":"python","codeSampleType":"brute-force","text":"# Please note that commented code is of brute force way but this fails for big numbers\n# def getSum(a: int, b: int) -> int:\n# # Loop until b becomes zero\n# while b != 0:\n# # If b is positive, increase a and decrease b\n# if b > 0:\n# a += 1 # simulate a + 1\n# b -= 1 # simulate b - 1\n# # If b is negative, decrease a and increase b\n# else:\n# a -= 1 # simulate a - 1\n# b += 1 # simulate b + 1\n\n# # Final result stored in a\n# return a\n# smater way of brute force\ndef getSum(a: int, b: int) -> int:\n MASK = 0xFFFFFFFF\n MAX = 0x7FFFFFFF\n\n while b != 0:\n carry = (a & b) & MASK\n a = (a ^ b) & MASK\n b = (carry << 1) & MASK\n\n return a if a <= MAX else ~(a ^ MASK)\n\n","description":"

\n \n The brute force approach calculates the sum of two integers by repeatedly\n incrementing or decrementing one number until the other becomes zero.\n This simulates addition using only basic operations and control flow.\n

\nSome test cases may fail due to timing or range issue for brute force case\n\n

\n Instead of using + or -, the algorithm adjusts\n the values step by step. If one number is positive, it increases the other.\n If one number is negative, it decreases the other.\n

\n\n

\n Although this method works conceptually, it becomes extremely slow when\n the numbers are large, making it unsuitable for real-world usage.\n

\n\n

\n Intuition (Simple Language):
\n Imagine you want to add 5 and 3, but you are not allowed to add directly.\n

\n\n

\n You can say: “I will add 1 to 5, three times.”\n After doing this step by step, you finally reach 8.\n

\n\n

\n The same idea applies for negative numbers — instead of adding,\n you subtract 1 repeatedly.\n

\n\n

\n This feels natural to understand, but very slow to execute.\n

\n\n

\n Steps to Solve (Brute Force):\n

\n\n

Check if b is zero — if yes, return a.
If b is positive, increment a and decrement b.
If b is negative, decrement a and increment b.
Repeat the process until b becomes zero.
Return a as the final sum.

\n\n

\n Time Complexity: O(|b|)
\n Space Complexity: O(1)
\n
\n\n \n Learn Complexity\n \n
\n
\n

\n"},{"@context":"https://schema.org","@type":"SoftwareSourceCode","programmingLanguage":"java","codeSampleType":"optimize","text":"public static class Solution {\n public int getSum(int a, int b) {\n // Continue until there is no carry\n while (b != 0) {\n // Carry is created where both bits are 1 (1 & 1 = 1, otherwise 0)\n // 'carry' stores the common set bits of 'a' and 'b', representing the carry bits.\n int carry = a & b;\n\n // XOR computes the sum without considering the carry (0+0=0, 0+1=1, 1+0=1, 1+1=0)\n // 'a' is updated with the sum of 'a' and 'b' without considering carry.\n a = a ^ b;\n\n // Shift the carry left by one bit to add it to the next higher bit position\n // 'carry << 1' shifts the carry bits one position to the left, preparing them to be added in the next iteration.\n b = carry << 1;\n }\n\n // When 'b' becomes 0, there's no more carry, so 'a' contains the final sum\n // After the loop finishes (no more carry), 'a' holds the final sum.\n return a;\n }\n}\n","description":"

\n\n

\n Binary addition can be split into two parts: calculating the sum without\n carry and calculating the carry itself. These two parts are combined\n repeatedly until no carry remains.\n

\n\n

\n By looping until the carry becomes zero, this method guarantees that all\n bits are added correctly, producing the final sum efficiently.\n

\n\n

\n Intuition (Simple Language):
\n Think about how you add numbers on paper.\n

\n\n

\n You write the number and carry extra value to the next column.\n Binary addition works the same way.\n

\n\n

\n XOR (^) gives the sum without carry, and\n AND (&) tells us where the carry is created.\n

\n\n

\n The carry belongs to the next higher bit, so we shift it left.\n We keep repeating this until there is no carry left.\n

\n\n

\n When no carry remains, the value left is the final answer.\n

\n\n

\n Steps to Solve:\n

\n\n

Start with two integers a and b.
Calculate the carry using a & b.
Calculate the sum without carry using a ^ b.
Shift the carry left by one bit using << 1.
Assign the XOR result back to a.
Assign the shifted carry back to b.
Repeat until b becomes zero.
Return a as the final sum.

\n\n

\n Time Complexity: O(1)
\n Space Complexity: O(1)
\n
\n\n \n Learn Complexity\n \n
\n
\n

\n"},{"@context":"https://schema.org","@type":"SoftwareSourceCode","programmingLanguage":"java","codeSampleType":"starter","text":"public static class Solution {\n \n public int getSum(int a, int b) {\n // TODO: Implement\n return 0;\n }\n}\n","description":"Sum of Two Integers implementation in java"},{"@context":"https://schema.org","@type":"SoftwareSourceCode","programmingLanguage":"java","codeSampleType":"brute-force","text":"public int getSum(int a, int b) {\n // Continue until b becomes zero\n while (b != 0) {\n // If b is positive, increment a and decrement b\n if (b > 0) {\n a++; // simulate a + 1\n b--; // simulate b - 1\n }\n // If b is negative, decrement a and increment b\n else {\n a--; // simulate a - 1\n b++; // simulate b + 1\n }\n }\n\n // a now contains the final sum\n return a;\n}\n","description":"

\nSome test cases may fail due to timing or range issue for brute force case\n\n

\n Instead of using + or -, the algorithm adjusts\n the values step by step. If one number is positive, it increases the other.\n If one number is negative, it decreases the other.\n

\n\n

\n Although this method works conceptually, it becomes extremely slow when\n the numbers are large, making it unsuitable for real-world usage.\n

\n\n

\n Intuition (Simple Language):
\n Imagine you want to add 5 and 3, but you are not allowed to add directly.\n

\n\n

\n You can say: “I will add 1 to 5, three times.”\n After doing this step by step, you finally reach 8.\n

\n\n

\n The same idea applies for negative numbers — instead of adding,\n you subtract 1 repeatedly.\n

\n\n

\n This feels natural to understand, but very slow to execute.\n

\n\n

\n Steps to Solve (Brute Force):\n

\n\n

Check if b is zero — if yes, return a.
If b is positive, increment a and decrement b.
If b is negative, decrement a and increment b.
Repeat the process until b becomes zero.
Return a as the final sum.

\n\n

\n Time Complexity: O(|b|)
\n Space Complexity: O(1)
\n
\n\n \n Learn Complexity\n \n
\n
\n

\n"},{"@context":"https://schema.org","@type":"SoftwareSourceCode","programmingLanguage":"cpp","codeSampleType":"optimize","text":"class Solution {\npublic:\n int getSum(int a, int b) {\n // Loop until no carry remains. This is the core of the bitwise addition.\n while (b != 0) {\n // Carry is generated when both bits are 1. The '&' (AND) operation identifies these bits.\n int carry = a & b;\n\n // XOR gives the sum without carry. The '^' (XOR) operation simulates addition without considering carry.\n a = a ^ b;\n\n // Shift carry left to add it to the higher bit. The '<< 1' operation moves the carry one position to the left.\n b = carry << 1;\n }\n\n // 'a' holds the final sum after all carries have been processed.\n return a;\n }\n};\n","description":"

\n\n

\n Binary addition can be split into two parts: calculating the sum without\n carry and calculating the carry itself. These two parts are combined\n repeatedly until no carry remains.\n

\n\n

\n By looping until the carry becomes zero, this method guarantees that all\n bits are added correctly, producing the final sum efficiently.\n

\n\n

\n Intuition (Simple Language):
\n Think about how you add numbers on paper.\n

\n\n

\n You write the number and carry extra value to the next column.\n Binary addition works the same way.\n

\n\n

\n XOR (^) gives the sum without carry, and\n AND (&) tells us where the carry is created.\n

\n\n

\n The carry belongs to the next higher bit, so we shift it left.\n We keep repeating this until there is no carry left.\n

\n\n

\n When no carry remains, the value left is the final answer.\n

\n\n

\n Steps to Solve:\n

\n\n

Start with two integers a and b.
Calculate the carry using a & b.
Calculate the sum without carry using a ^ b.
Shift the carry left by one bit using << 1.
Assign the XOR result back to a.
Assign the shifted carry back to b.
Repeat until b becomes zero.
Return a as the final sum.

\n\n

\n Time Complexity: O(1)
\n Space Complexity: O(1)
\n
\n\n \n Learn Complexity\n \n
\n
\n

\n"},{"@context":"https://schema.org","@type":"SoftwareSourceCode","programmingLanguage":"cpp","codeSampleType":"starter","text":"class Solution {\npublic:\n /**\n * Calculates the sum of two integers without using the + and - operators.\n * \n * @param a The first integer.\n * @param b The second integer.\n * @return The sum of a and b.\n */\n int getSum(int a, int b) {\n // TODO: Implement bitwise addition logic here.\n return 0; // Placeholder return value.\n }\n};\n","description":"Sum of Two Integers implementation in cpp"},{"@context":"https://schema.org","@type":"SoftwareSourceCode","programmingLanguage":"cpp","codeSampleType":"brute-force","text":"int getSum(int a, int b) {\n // Loop until b becomes zero\n while (b != 0) {\n // If b is positive, increment a and decrement b\n if (b > 0) {\n a++; // simulate a + 1\n b--; // simulate b - 1\n }\n // If b is negative, decrement a and increment b\n else {\n a--; // simulate a - 1\n b++; // simulate b + 1\n }\n }\n\n // Final sum stored in a\n return a;\n}\n","description":"

\nSome test cases may fail due to timing or range issue for brute force case\n\n

\n Instead of using + or -, the algorithm adjusts\n the values step by step. If one number is positive, it increases the other.\n If one number is negative, it decreases the other.\n

\n\n

\n Although this method works conceptually, it becomes extremely slow when\n the numbers are large, making it unsuitable for real-world usage.\n

\n\n

\n Intuition (Simple Language):
\n Imagine you want to add 5 and 3, but you are not allowed to add directly.\n

\n\n

\n You can say: “I will add 1 to 5, three times.”\n After doing this step by step, you finally reach 8.\n

\n\n

\n The same idea applies for negative numbers — instead of adding,\n you subtract 1 repeatedly.\n

\n\n

\n This feels natural to understand, but very slow to execute.\n

\n\n

\n Steps to Solve (Brute Force):\n

\n\n

Check if b is zero — if yes, return a.
If b is positive, increment a and decrement b.
If b is negative, decrement a and increment b.
Repeat the process until b becomes zero.
Return a as the final sum.

\n\n

\n Time Complexity: O(|b|)
\n Space Complexity: O(1)
\n
\n\n \n Learn Complexity\n \n
\n
\n

\n"},{"@context":"https://schema.org","@type":"SoftwareSourceCode","programmingLanguage":"TypeScript","codeSampleType":"optimize","text":"function getSum(a: number, b: number): number {\n // Continue until there is no carry left\n while (b !== 0) {\n // Carry is generated where both bits are 1\n // The '&' (AND) operation identifies bits that are 1 in both 'a' and 'b'.\n // This represents the carry that will be added to the next higher bit.\n const carry = a & b;\n\n // XOR gives sum without carry\n // The '^' (XOR) operation gives the sum of 'a' and 'b' without considering the carry.\n // It effectively adds the bits where only one of 'a' or 'b' is 1.\n a = a ^ b;\n\n // Shift carry left to add it in the next higher bit\n // The '<< 1' operation shifts the carry one bit to the left.\n // This prepares the carry to be added to the next higher bit in the next iteration.\n b = carry << 1;\n }\n\n // When no carry remains, 'a' contains the final sum\n // After the loop completes, 'a' holds the final sum of the two numbers.\n return a;\n}\n","description":"Sum of Two Integers implementation in TypeScript"},{"@context":"https://schema.org","@type":"SoftwareSourceCode","programmingLanguage":"TypeScript","codeSampleType":"starter","text":"\nfunction getSum(a: number, b: number): number {\n // TODO: Implement the function to calculate the sum of two integers without using the + and - operators.\n // You can use bitwise operators like &, |, ^, ~, <<, and >> to achieve this.\n return 0; // Placeholder return value\n};\n","description":"Sum of Two Integers implementation in TypeScript"},{"@context":"https://schema.org","@type":"SoftwareSourceCode","programmingLanguage":"TypeScript","codeSampleType":"brute-force","text":"function getSum(a: number, b: number): number {\n // Continue until b becomes zero\n while (b !== 0) {\n // If b is positive, increment a and decrement b\n if (b > 0) {\n a++; // simulate a + 1\n b--; // simulate b - 1\n }\n // If b is negative, decrement a and increment b\n else {\n a--; // simulate a - 1\n b++; // simulate b + 1\n }\n }\n\n // a now contains the final sum\n return a;\n}","description":"

\nSome test cases may fail due to timing or range issue for brute force case\n\n

\n Instead of using + or -, the algorithm adjusts\n the values step by step. If one number is positive, it increases the other.\n If one number is negative, it decreases the other.\n

\n\n

\n Although this method works conceptually, it becomes extremely slow when\n the numbers are large, making it unsuitable for real-world usage.\n

\n\n

\n Intuition (Simple Language):
\n Imagine you want to add 5 and 3, but you are not allowed to add directly.\n

\n\n

\n You can say: “I will add 1 to 5, three times.”\n After doing this step by step, you finally reach 8.\n

\n\n

\n The same idea applies for negative numbers — instead of adding,\n you subtract 1 repeatedly.\n

\n\n

\n This feels natural to understand, but very slow to execute.\n

\n\n

\n Steps to Solve (Brute Force):\n

\n\n

Check if b is zero — if yes, return a.
If b is positive, increment a and decrement b.
If b is negative, decrement a and increment b.
Repeat the process until b becomes zero.
Return a as the final sum.

\n\n

\n Time Complexity: O(|b|)
\n Space Complexity: O(1)
\n
\n\n \n Learn Complexity\n \n
\n
\n

\n"}]}

78. Sum of Two Integers

Understand Before Coding

Visualize This Problem

Draw Your Approach

Example 1:

Example 2:

Constraints:

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78. Sum of Two Integers

Understand Before Coding

Visualize This Problem

Draw Your Approach

Example 1:

Example 2:

Constraints:

Learn

Topics

Companies

Hint

Use Cases