Find an element from a array - c++

Question

I have an array which spans as follows. So the example to understand is consider a block with 6 faces. The array contains the index in this array from which the faces for a particular colour will start.
array[0] 0
array[1] 2
array[2] 4
array[3] 5
so this means that colour 0 is for faces 0 and 1, colour 1 is for faces 2 and 3,
and colour 2 is only for face 4
But the array will not always look like this. if there is a block with just a single colour the array would look like
array[0] 0
array[1] 1
which means face 0 is coloured with 0 color, and faces 1,2,3,4,5 are coloured with color 1
I will get an input as the face number and need to find the corresponding colour
I tried it with a for loop
for (int index = 0; index < array.size(); ++index)
{
if (array[index] == input)
{
return index;
}
if (array[index] < input)
{
return index - 1;
}
}
but the answer is not always correct . Can this be done with while . Please help

Answer 1

As I understand, you want to find the largest index of the array which is less or equal to the given input.
Then you can use the following with binary search.
std::size_t getIndexFor(const std::vector<int>& v, int input)
{
auto it = std::lower_bound(v.begin(), v.end(), input);
if (it != v.begin() && (it == v.end() || *it != input)) {
--it;
}
return std::distance(v.begin(), it);
}
Demo
A simpler (and linear) way:
std::size_t getIndexFor(const std::vector<int>& v, int input)
{
auto it = std::find_if(v.begin(), v.end(), [&](int e){ return e > input;});
if (it == v.begin()) {
throw std::runtime_error("input not handled");
}
--it;
return std::distance(v.begin(), it);
}
Demo

Answer 2

You would just do the same thing, looping through the array to search for the term you want. Of course if it's a sorted array this would be much faster, so something similar to prehaps:
for(int i = 0; i < arraySize; i++){
if(array[i] == itemToFind){
break;
}
}

Answer 3

You are very close to the right solution, just change array[index] < input into array[index] > input:
#include <iostream>
#include <vector>
std::vector<int> array = {0, 2, 4, 5};
int subs(int input)
{
for (int index = 0; index < array.size(); ++index)
{
if (array[index] == input)
{
return index;
}
if (array[index] > input)
{
return index - 1;
}
}
return -1;
}
int main(int argc, const char *argv[])
{
for (int i = 0; i <= 5; i++) {
std::cout<<"input: "<<i<<" "<<"color: "<<subs(i)<<std::endl;
}
return 0;
}
This gives:
input: 0 color: 0
input: 1 color: 0
input: 2 color: 1
input: 3 color: 1
input: 4 color: 2
input: 5 color: 3

Answer 4

I used the for loop also but using break after one if condition that checks if the input equals any element in the array.
#include <iostream>
using namespace std;
int main(){
string arr = " ";
int x;
cout << "Enter your input";
cin >> x;
for (int i =0; i< arr.length(); i++ ){
if (arr[i] == x){
break;
}
}
return 0;
}

Answer 5

Another, maybe simpler solution:
int findIndex( const std::vector<int>& vec, int input )
{
int count = vec.size()-1;
for ( auto it = vec.rbegin(); it != vec.rend(); it++)
{
if ( *it<=input )
{
return count;
}
count--;
}
return -1; // if nothing found
}
int main()
{
std::vector<int> vec { 0,2,4,5 };
for ( int input=0; input <=5; input++)
{
std::cout << findIndex( vec, input ) << " " << input << std::endl;
}
}
Output:
0 0
0 1
1 2
1 3
2 4
3 5