---
title: "3SUM"
chunk: 2/3
source: "https://en.wikipedia.org/wiki/3SUM"
category: "reference"
tags: "science, encyclopedia"
date_saved: "2026-05-05T11:01:54.553179+00:00"
instance: "kb-cron"
---

=== Three different arrays ===
Instead of searching for the 3 numbers in a single array, we can search for them in 3 different arrays. I.e., given three arrays X, Y and Z, find three numbers a∈X, b∈Y, c∈Z, such that ⁠
  
    
      
        a
        +
        b
        +
        c
        =
        0
      
    
    {\displaystyle a+b+c=0}
  
⁠. Call the 1-array variant 3SUM×1 and the 3-array variant 3SUM×3.
Given a solver for 3SUM×1, the 3SUM×3 problem can be solved in the following way (assuming all elements are integers): 

For every element in X, Y and Z, set: ⁠
  
    
      
        X
        [
        i
        ]
        ←
        X
        [
        i
        ]
        ∗
        10
        +
        1
      
    
    {\displaystyle X[i]\gets X[i]*10+1}
  
⁠, ⁠
  
    
      
        Y
        [
        i
        ]
        ←
        Y
        [
        i
        ]
        ∗
        10
        +
        2
      
    
    {\displaystyle Y[i]\gets Y[i]*10+2}
  
⁠, ⁠
  
    
      
        Z
        [
        i
        ]
        ←
        Z
        [
        i
        ]
        ∗
        10
        −
        3
      
    
    {\displaystyle Z[i]\gets Z[i]*10-3}
  
⁠.
Let S be a concatenation of the arrays X, Y and Z.
Use the 3SUM×1 oracle to find three elements ⁠
  
    
      
        
          a
          ′
        
        ∈
        S
        ,
         
        
          b
          ′
        
        ∈
        S
        ,
         
        
          c
          ′
        
        ∈
        S
      
    
    {\displaystyle a'\in S,\ b'\in S,\ c'\in S}
  
⁠ such that ⁠
  
    
      
        
          a
          ′
        
        +
        
          b
          ′
        
        +
        
          c
          ′
        
        =
        0
      
    
    {\displaystyle a'+b'+c'=0}
  
⁠.
Return ⁠
  
    
      
        a
        ←
        (
        
          a
          ′
        
        −
        1
        )
        
          /
        
        10
        ,
         
        b
        ←
        (
        
          b
          ′
        
        −
        2
        )
        
          /
        
        10
        ,
         
        c
        ←
        (
        
          c
          ′
        
        +
        3
        )
        
          /
        
        10
      
    
    {\displaystyle a\gets (a'-1)/10,\ b\gets (b'-2)/10,\ c\gets (c'+3)/10}
  
⁠.
By the way we transformed the arrays, it is guaranteed that a∈X, b∈Y, c∈Z.

=== Convolution sum ===
Instead of looking for arbitrary elements of the array such that:

  
    
      
        S
        [
        k
        ]
        =
        S
        [
        i
        ]
        +
        S
        [
        j
        ]
      
    
    {\displaystyle S[k]=S[i]+S[j]}
  

the convolution 3sum problem (Conv3SUM) looks for elements in specific locations:

  
    
      
        S
        [
        i
        +
        j
        ]
        =
        S
        [
        i
        ]
        +
        S
        [
        j
        ]
      
    
    {\displaystyle S[i+j]=S[i]+S[j]}
  

==== Reduction from Conv3SUM to 3SUM ====
Given a solver for 3SUM, the Conv3SUM problem can be solved in the following way.

Define a new array T, such that for every index i: 
  
    
      
        T
        [
        i
        ]
        =
        2
        n
        S
        [
        i
        ]
        +
        i
      
    
    {\displaystyle T[i]=2nS[i]+i}
  
 (where n is the number of elements in the array, and the indices run from 0 to n-1).
Solve 3SUM on the array T.
Correctness proof:

If in the original array there is a triple with 
  
    
      
        S
        [
        i
        +
        j
        ]
        =
        S
        [
        i
        ]
        +
        S
        [
        j
        ]
      
    
    {\displaystyle S[i+j]=S[i]+S[j]}
  
, then 
  
    
      
        T
        [
        i
        +
        j
        ]
        =
        2
        n
        S
        [
        i
        +
        j
        ]
        +
        i
        +
        j
        =
        (
        2
        n
        S
        [
        i
        ]
        +
        i
        )
        +
        (
        2
        n
        S
        [
        j
        ]
        +
        j
        )
        =
        T
        [
        i
        ]
        +
        T
        [
        j
        ]
      
    
    {\displaystyle T[i+j]=2nS[i+j]+i+j=(2nS[i]+i)+(2nS[j]+j)=T[i]+T[j]}
  
, so this solution will be found by 3SUM on T.
Conversely, if in the new array there is a triple with 
  
    
      
        T
        [
        k
        ]
        =
        T
        [
        i
        ]
        +
        T
        [
        j
        ]
      
    
    {\displaystyle T[k]=T[i]+T[j]}
  
, then 
  
    
      
        2
        n
        S
        [
        k
        ]
        +
        k
        =
        2
        n
        (
        S
        [
        i
        ]
        +
        S
        [
        j
        ]
        )
        +
        (
        i
        +
        j
        )
      
    
    {\displaystyle 2nS[k]+k=2n(S[i]+S[j])+(i+j)}
  
. Because 
  
    
      
        i
        +
        j
        <
        2
        n
      
    
    {\displaystyle i+j<2n}
  
, necessarily 
  
    
      
        S
        [
        k
        ]
        =
        S
        [
        i
        ]
        +
        S
        [
        j
        ]
      
    
    {\displaystyle S[k]=S[i]+S[j]}
  
 and 
  
    
      
        k
        =
        i
        +
        j
      
    
    {\displaystyle k=i+j}
  
, so this is a valid solution for Conv3SUM on S.

==== Reduction from 3SUM to Conv3SUM ====
Given a solver for Conv3SUM, the 3SUM problem can be solved in the following way.
The reduction uses a hash function. As a first approximation, assume that we have a linear hash function, i.e. a function h such that:

  
    
      
        h
        (
        x
        +
        y
        )
        =
        h
        (
        x
        )
        +
        h
        (
        y
        )
      
    
    {\displaystyle h(x+y)=h(x)+h(y)}
  

Suppose that all elements are integers in the range: 0...N−1, and that the function h maps each element to an element in the smaller range of indices: 0...n−1. Create a new array T and send each element of S to its hash value in T, i.e., for every x in S(⁠
  
    
      
        ∀
        x
        ∈
        S
      
    
    {\displaystyle \forall x\in S}
  
⁠):

  
    
      
        T
        [
        h
        (
        x
        )
        ]
        =
        x
      
    
    {\displaystyle T[h(x)]=x}
  

Initially, suppose that the mappings are unique (i.e. each cell in T accepts only a single element from S). Solve Conv3SUM on T. Now: