How can I get the maximum or minimum value in a vector?

Question

How can I get the maximum or minimum value in a vector in C++?

And am I wrong in assuming it would be more or less the same with an array?

I need an iterator, right? I tried it with max_element, but I kept getting an error.

vector<int>::const_iterator it;
it = max_element(cloud.begin(), cloud.end());

error: request for member ‘begin’ in ‘cloud’, which is of non-class type ‘int [10]’

Answer 1

If you want to use the function std::max_element() from <algorithm>, the way you have to do it is:

#include <algorithm>

double max = *std::max_element(vector.begin(), vector.end());
std::cout << "Max value: " << max << std::endl;

Answer 2

_{Answer on the behalf of the author}

for (unsigned int i = 0; i < cdf.size(); i++)
  if (cdf[i] < cdfMin)
    cdfMin = cdf[i];

where cdf is a vector.

Answer 3

You can print it directly using the max_element or min_element function.

For example:

cout << *max_element(v.begin(), v.end());

cout << *min_element(v.begin(), v.end());

Answer 4

In C++11, you can use some function like that:

int maxAt(std::vector<int>& vector_name) {
    int max = INT_MIN;
    for (auto val : vector_name) {
         if (max < val) max = val;
    }
    return max;
}

Answer 5

Using C++11/C++0x compile flags, you can

auto it = max_element(std::begin(cloud), std::end(cloud)); // C++11

Otherwise, write your own:

template <typename T, size_t N> const T* mybegin(const T (&a)[N]) { return a; }
template <typename T, size_t N> const T* myend  (const T (&a)[N]) { return a+N; }

See it live at http://ideone.com/aDkhW:

#include <iostream>
#include <algorithm>

template <typename T, size_t N> const T* mybegin(const T (&a)[N]) { return a; }
template <typename T, size_t N> const T* myend  (const T (&a)[N]) { return a+N; }

int main()
{
    const int cloud[] = { 1,2,3,4,-7,999,5,6 };

    std::cout << *std::max_element(mybegin(cloud), myend(cloud)) << '\n';
    std::cout << *std::min_element(mybegin(cloud), myend(cloud)) << '\n';
}

Oh, and use std::minmax_element(...) if you need both at once :/

Answer 6

Just this:

// assuming "cloud" is:
// int cloud[10]; 
// or any other fixed size

#define countof(x) (sizeof(x)/sizeof((x)[0]))

int* pMax = std::max_element(cloud, cloud + countof(cloud));

Answer 7

Let,

 #include <vector>

 vector<int> v {1, 2, 3, -1, -2, -3};

If the vector is sorted in ascending or descending order then you can find it with complexity O(1).

For a vector of ascending order the first element is the smallest element, you can get it by v[0] (0 based indexing) and last element is the largest element, you can get it by v[sizeOfVector-1].

If the vector is sorted in descending order then the last element is the smallest element,you can get it by v[sizeOfVector-1] and first element is the largest element, you can get it by v[0].

If the vector is not sorted then you have to iterate over the vector to get the smallest/largest element.In this case time complexity is O(n), here n is the size of vector.

int smallest_element = v[0]; //let, first element is the smallest one
int largest_element = v[0]; //also let, first element is the biggest one
for(int i = 1; i < v.size(); i++)  //start iterating from the second element
{
    if(v[i] < smallest_element)
    {
       smallest_element = v[i];
    }
    if(v[i] > largest_element)
    {
       largest_element = v[i];
    }
}

You can use iterator,

for (vector<int>:: iterator it = v.begin(); it != v.end(); it++)
{
    if(*it < smallest_element) //used *it (with asterisk), because it's an iterator
    {
      smallest_element = *it;
    }
    if(*it > largest_element)
    {
      largest_element = *it;
    }
}

You can calculate it in input section (when you have to find smallest or largest element from a given vector)

int smallest_element, largest_element, value;
vector <int> v;
int n;//n is the number of elements to enter
cin >> n;
for(int i = 0;i<n;i++)
{
    cin>>value;
    if(i==0)
    {
        smallest_element= value; //smallest_element=v[0];
        largest_element= value; //also, largest_element = v[0]
    }

    if(value<smallest_element and i>0)
    {
        smallest_element = value;
    }

    if(value>largest_element and i>0)
    {
        largest_element = value;
    }
    v.push_back(value);
}

Also you can get smallest/largest element by built in functions

#include<algorithm>

int smallest_element = *min_element(v.begin(),v.end());

int largest_element  = *max_element(v.begin(),v.end());

You can get smallest/largest element of any range by using this functions. such as,

vector<int> v {1,2,3,-1,-2,-3};

cout << *min_element(v.begin(), v.begin() + 3); //this will print 1,smallest element of first three elements

cout << *max_element(v.begin(), v.begin() + 3); //largest element of first three elements

cout << *min_element(v.begin() + 2, v.begin() + 5); // -2, smallest element between third and fifth element (inclusive)

cout << *max_element(v.begin() + 2, v.begin()+5); //largest element between third and first element (inclusive)

I have used asterisk (*), before min_element()/max_element() functions. Because both of them return iterator. All codes are in c++.

Answer 8

You can use max_element to get the maximum value in vector. The max_element returns an iterator to largest value in the range, or last if the range is empty. As an iterator is like pointers (or you can say pointer is a form of iterator), you can use a * before it to get the value. So as per the problem you can get the maximum element in an vector as:

int max=*max_element(cloud.begin(), cloud.end());

It will give you the maximum element in your vector "cloud". Hope it helps.

Answer 9

If you want to use an iterator, you can do a placement-new with an array.

std::array<int, 10> icloud = new (cloud) std::array<int,10>;

Note the lack of a () at the end, that is important. This creates an array class that uses that memory as its storage, and has STL features like iterators.

(This is C++ TR1/C++11 by the way)

Answer 10

Assuming cloud is int cloud[10] you can do it like this: int *p = max_element(cloud, cloud + 10);