One simple yet effective method entails examining each element in the array and searching through the remaining elements for its pair to obtain the desired result when added together.
For instance, consider the following input:
arr = [1,4,2,3,0,5,4,3]
res = 7
Initially, the number 1 (the 0th element) is evaluated. Subsequently, each subsequent element (ranging from 4 to 3) is added to it and compared against 7. In this case, there is no occurrence of 6 in the array, resulting in no match found.
Moving on to the evaluation of the number 4, we encounter some interesting findings:
4 + 2 = 6 // not a pair
4 + 3 = 7 // a pair found!
4 + 0 = 4 // not a pair
4 + 5 = 9 // not a pair
4 + 4 = 8 // not a pair
4 + 3 = 7 // another pair found!
Through this search process, two sets of pairs with identical elements - [4, 3] are identified. Moving past the obvious (2, 5 pair), the examination then shifts to 3:
3 + 0 = 3; // -
3 + 5 = 8; // -
3 + 4 = 7; // +
3 + 3 = 6; // -
An observable pattern emerges during this iteration: the current evaluation concludes by identifying the pair of first 3, second 4. Both choices appear to align with the specified criteria.
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The alternative approach introduces an unconventional perspective. The author's interpretation considers indexOf as a straightforward operation with constant time complexity (which is not accurate) and overlooks a crucial distinction: indexOf inherently targets the first element adhering to its predicate.
As exemplified, during the execution of the reduce function over the initial 4 in the array, indexOf verifies the presence of
3</code - leading to the discovery of the first <code>3
encountered:
// reduce on b = 4, index = 1
search = 7 - 4; // 3
searchIndex = arr.indexOf(3); // 3
However, this particular pairing resurfaces when evaluating the first occurrence of 3</code (as <code>indexOf also identifies the first 4).
// reduce on b = 3, index = 3
search = 7 - 3; // 4
searchIndex = arr.indexOf(4); // 1
Intriguingly, despite the redundancy within this set, it unintentionally compensates for neglecting to reevaluate the 'first instances' with respect to both the subsequent 4 and 3.
// reduce on b = 4, index = 6
search = 7 - 4; // 3
searchIndex = arr.indexOf(3); // 3
// reduce on b = 3, index = 7
search = 7 - 3; // 4
searchIndex = arr.indexOf(4); // 1
Notably, further pairs are identified while one remains unnoticed: the last elements of the array lack corresponding pairs.
Can the secondary approach be rectified? Certainly, a slight enhancement involves providing the secondary argument into indexOf:
let searchIndex = pairArr.indexOf(search, index + 1);
...however, this adjustment solely caters to arrays comprising unique elements. Otherwise, resorting to an explicit loop would essentially transform this strategy into a far more complex rendition of the initial one.