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LESSON 265 OF 667 · PAIRS, SORTING & BUILT-INS

Min, Max and Sum in C++: min_element, max_element, accumulate

Lesson Read the full explanation

TL ; DR:

*min_element(v.begin(), v.end()) gives the smallest value of a vector.

*max_element gives the largest value.

accumulate adds all the values up.

vector<int> prices = {4, 9, 2, 7};

cout << *min_element(prices.begin(), prices.end()) << endl;    // 2

cout << *max_element(prices.begin(), prices.end()) << endl;    // 9

int total = accumulate(prices.begin(), prices.end(), 0);

cout << total << endl;                                          // 22

Full lesson:

In Maximum Number In Array, a loop found the largest value. In Sum of Numbers, a loop added numbers up. These jobs come up all the time. The C++ library has them ready-made.

min_element(v.begin(), v.end()) finds the smallest value of the vector v.
max_element(v.begin(), v.end()) finds the largest value.
accumulate(v.begin(), v.end(), 0) adds all the values to a start value, 0.

begin() and end() mark the whole vector. min_element returns where the smallest value is, not the value. The * in front reads the value there.

The three come from the headers <algorithm> and <numeric>. This lesson already includes them.

The start value also sets the total's type. With 0 the total is an int. For doubles, start at 0.0, or the fractions get lost.

max(a, b) from Maximum Number In Array compares two values. max_element looks at a whole vector.

Watch out for an empty vector. There, min_element returns v.end(), a spot past the last value. Reading it with * is undefined behavior. The program may crash or print garbage. Check v.empty() first. accumulate on an empty vector gives its start value, 0.

Your task

minutes holds how many minutes you practiced coding on each day of a week.

Store the shortest day in shortest and the longest day in longest. Store the week's total in total.

Use the built-ins from this lesson. The code at the bottom prints all three.

Expected output:

shortest = 15
longest = 60
total = 235

STEP 1 OF 3

Guided

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C++ · PAIRS, SORTING & BUILT-INS

Min Max Sum in C++, explained

Ready-made helpers for common loops

Finding the smallest value, the largest value and the total of a vector are jobs a loop can do, and they come up constantly. The C++ library has them ready-made:

min_element(v.begin(), v.end()) finds the smallest value of the vector v.
max_element(v.begin(), v.end()) finds the largest value.
accumulate(v.begin(), v.end(), 0) adds all the values to a start value, here 0.

begin() and end() mark the whole vector. min_element and max_element live in <algorithm> and accumulate in <numeric>. The lesson already includes both headers.

vector<int> prices = {4, 9, 2, 7};
cout << *min_element(prices.begin(), prices.end()) << endl;   // 2
cout << *max_element(prices.begin(), prices.end()) << endl;   // 9
cout << accumulate(prices.begin(), prices.end(), 0) << endl;  // 22

Why the star?

min_element doesn't return the smallest value. It returns an iterator, which marks where that value sits in the vector. The * in front reads the value at that spot. Without it the code doesn't compile where an int is expected. Because the result is a position, you can also turn it into an index by subtracting begin(): for the minutes in this lesson, min_element(...) - minutes.begin() is 2, the index of the 15.

The start value sets the type

accumulate adds into a variable that has the type of the start value. With 0 that is an int. For a vector of doubles this loses the fractions: with {1.5, 2.5, 3.5}, a start of 0 gives 6, while a start of 0.0 gives 7.5. Use 0.0 for doubles.

Watch out for an empty vector

On an empty vector, min_element returns v.end(), a spot past the last value, where there is nothing to read. Reading it with * is undefined behavior: the program may crash or print garbage. Check v.empty() first. accumulate on an empty vector is safe: it returns the start value, 0.

Built-ins against loops

You can always write the loop yourself, and a loop is the right tool when the rule is unusual. The built-ins are shorter, say what they do in their name, and have no starting-value mistake to make: a hand-written minimum that starts at 0 gives a wrong answer for a vector of positive numbers. For a plain smallest, largest or total, prefer them.

The built-ins work on a range, so they can also cover part of a vector. For example, accumulate(v.begin(), v.begin() + 3, 0) adds only the first three values.

The exercise, step by step

minutes holds the minutes practiced on each day of a week. The task stores the shortest day, the longest day and the week's total.

Shortest: int shortest = *min_element(minutes.begin(), minutes.end());
Longest: the same with max_element.
Total: accumulate over the whole vector with a start value of 0, which also makes the total an int.

vector<int> minutes = {25, 40, 15, 60, 30, 45, 20};

int shortest = *min_element(minutes.begin(), minutes.end());
int longest = *max_element(minutes.begin(), minutes.end());
int total = accumulate(minutes.begin(), minutes.end(), 0);

cout << "shortest = " << shortest << endl;
cout << "longest = " << longest << endl;
cout << "total = " << total << endl;

The output:

shortest = 15
longest = 60
total = 235

Both min_element and max_element return the first matching position when a value appears more than once, which only matters if you use the position and not just the value.

Key takeaways

*min_element(v.begin(), v.end()) and *max_element(...) give the smallest and largest values.
accumulate(v.begin(), v.end(), 0) sums a vector; the start value decides the type, so use 0.0 for doubles.
Never dereference the result of min_element or max_element on an empty vector.

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