array - efficiency & general approach towards the question - eclipse

Question

Edit
**After quite a bit of playing around with my code I have written few versions, one of them is what I sought for.

Answer 1

I'd do the following:
Create a helper Set with all the numbers from 1 to 100 - O(n) in time and space
Create an (initially) empty Set to record the indexes with 0s - O(1)
Go over the array - O(n):
If the value is 0, add the index to the index set
If the value isn't 0, remove it from the helper Set
Go over the helper set, and assign the remaining values to the indexes saved to the index set - O(m), where m<=n
All in all - an O(n) solution in time and space.
In java:
int[] numbers = /* the array with missing numbers */
Set<Integer> allNumbers = IntStream.rangeClosed(1, 100).boxed().Collect(Collectors.toSet());
Set<Ineteger> missingIndexes = new HashSet<>();
for (int i = 0; i < numbers.length; ++i) {
if (numbers[i] == 0) {
missingIndexes.add(i);
} else {
allNumbers.remove(i);
}
}
Iterator<Integer> numberIter = allNumbers.iterator();
Iterator<Integer> indexIter = missingIndexes.iterator();
while (numberIter.hasNext() && indexIter.hasNext()) {
numbers[indexIter.next()] = numberIter.next();
}

Answer 2

If you can sacrifice a little more space to optimize for time.
Basically just create another missing array and fill it with what's missing by traversing your helper array.
Modified your original solution.
int input [] = {5,6,0,3,0,2,1};
int output [] = new int[input.length];
boolean [] helper = new boolean[input.length];
for(int i = 0; i <input.length; i++)
{
if(input[i] != 0)
helper[i] = true;
}
int missing [] = new int[input.length];
int missingCount = 0;
for(int j = 0; j < helper.length; j++)
{
if(!helper[j]){
missing[missingCount++] = j;
}
}
missingCount = 0;
for(int j = 0; j < input.length; j++){
if(input[j]==0){
input[j]=missing[missingCount++];
}
}

Answer 3

Below code can find the missing element and add them back in O(n) complexity:
notFound - Will store index or location of input array having number
with zero
numberDetails - Contains details of numbers whether its present in
input array or not(true or false)
Example: input[3] = false means 4 (3+1) is not present in input array
, input[4] = true means 5 (4+1) is present in input array
int input[] = { 5, 6, 0, 3, 0, 2, 1 };
int notFound[] = new int[input.length];
boolean[] numberDetails = new boolean[input.length];
int notFoundIndex=0;
for(int i=0;i<input.length;i++) {
if(input[i]==0) {
notFound[notFoundIndex++]=i;
}
else {
numberDetails[input[i]-1]=true;
}
}
notFoundIndex=0;
for(int j=0;j<numberDetails.length;j++) {
if(!numberDetails[j]) {
input[notFound[notFoundIndex++]] = j+1;
}
}
System.out.println(Arrays.toString(input));