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JavaScript: Simple function to find positive missing integers in Array of numbers.
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/* Write a function solution(A); that, given an array A of N integers, returns the smallest positive integer (greater than 0) that does not occur in A. | |
For example, given A = [1, 3, 6, 4, 1, 2], the function should return 5. | |
Given A = [1, 2, 3], the function should return 4. | |
Given A = [−1, −3], the function should return 1. | |
Write an efficient algorithm for the following assumptions: | |
N is an integer within the range [1..100,000]; | |
each element of array A is an integer within the range [−1,000,000..1,000,000]. */ | |
const A = [-1, -4]; | |
function solution(A) { | |
let max = A.reduce((acc, val) => acc > val ? acc : val); | |
let missing = 1; | |
for (let i = max + 1; i > 0; i--) { | |
found = A.find(el => el == i); | |
if(found === undefined) { | |
missing = i; | |
} | |
} | |
return missing; | |
} | |
console.log(solution(A)); |
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below come with 100%:
function solution(A){
let counter = 0;
A.sort(function (a, b) {
return a - b;
});
for (let i = 0; i < A.length; i++) {
const element = A[i];
if (element > 0) {
if (element != counter) {
counter++;
}
if (element == counter) {
continue;
}
return counter;
}
}
counter++;
return counter;
}