# Exercise 3-4

Write a program to report the length of the longest and shortest string in its input.

# Solution

I notice that the solution to this problem will be somewhat similar to the Solution to Exercise 3-3 (identify number of distinct elements in a vector). Except this time we are determining the length of the longest and shortest (string) elements. I therefore expect the eventual program to be similar.

## Strategy

- Ask user to provide the list of words so we can append the string elements to a vector,
**v**. - Compute the size of vector,
**N**. - If vector size is 0, output an error message (as we need at least 1 string). Exit program peacefully with return code 1.
- If vector size is 1, we compute the size of the string element
**v[**0**]**. The length of the longest string,**SL**, will be the same as the length of the shortest string,**SS**. - If vector size is 2 or above, we apply an algorithm to compute
**SL**and**SS**(to be explained further below).

## Algorithm

The algorithm (for the case of vector size of 2 or above) is described as followings:

- Notice that this time we do not need to sort the vector – the condition checks downstream will be sufficient to update
**SS**and**SL**. - We initialise
**SL**and**SS**to the size of the first string element**v[**0**]**.**SL**and**SS**will be updated accordingly during the comparison step. - We initialise the index
**B**to 1. This will be used for comparing the length of the**v[B]**with respect to**SS**and**SL**. - We perform
**N-1**number of comparisons between**v[B]**and the current**SL**and**SS**. (e.g. if there are 10 elements, we can only compare 9 sets of adjacent elements). - During each comparison step, if
**v[B].size()**is larger than**SL**, we re-assign**SL**to**v[B].size()**. - If
**v[B].size()**is smaller than**SS**, we re-assign**SS**to**v[B].size()**. - (If the two element lengths are equal, we do nothing.)
- We then increment the indices
**B**by 1 for the next comparisons. - We display the final value of
**SS**and**SL**– these are the lengths of the shortest and longest strings elements within the vector**v**.

## The Program

Putting this all together, we have our full program.

#include <iostream> #include <string> #include <vector> using std::cin; // <iostream> using std::cout; // <iostream> using std::endl; // <iostream> using std::string; // <string> using std::vector; // <string> int main() { // display header message cout << "***************************************************************\n" "*** This program reports the longest and shortest strings ***\n" "***************************************************************\n"; cout << endl; // ask for a list of numbers and store the list as a vector cout << "Enter a list of words one by one: "; vector<string> v; string x; while (cin >> x) v.push_back(x); // append new input to the vector cout << endl; // define and compute core vector variables typedef vector<string>::size_type vecSize; // define a type for vector size related variables vecSize N = v.size(); // number of elements in the vector vecSize numLoops = N - 1; // number of (comparison) operators required typedef string::size_type strSize; // define a type for string size related variables strSize SL; // the length of the longest word strSize SS; // the length of the shortest word // Check vector size, action accordingly if (N ==0 ) { cout << "You need to enter at least 1 word! " << endl; return 1; } else if (N ==1 ) { SL = v[0].size(); cout << "Only 1 string supplied. The length of string = " << SL << endl; return 0; } else { // display some results to console window cout << "Vector size (number of words entered): " << N << endl; cout << endl; // declare new variables vecSize A = 0; // vector index vecSize B = 1; // vector index SS = v[0].size(); // the length of the shortest word SL = v[0].size(); // the length of the longest word // Loop through the vector, compute ND, and compute SS and SL for (vecSize i = 0; i != numLoops; ++i) { if (v[B].size() > SL) { SL = v[B].size(); } if (v[B].size() < SS) { SS = v[B].size(); } ++B; } // Display final results cout << endl; cout << "Length of shortest word: " << SS << endl; cout << "Length of longest word: " << SL << endl; } return 0; }

# Result

Below shows the results of the 3 sets of test:

- N = 0
- N=1
- N>=2

The results appear to agree well with the algorithm used.

## N = 0

*************************************************************** *** This program reports the longest and shortest strings *** *************************************************************** Enter a list of words one by one: ^Z You need to enter at least 1 word! Process returned 1 (0x1) execution time : 3.156 s

## N = 1

*************************************************************** *** This program reports the longest and shortest strings *** *************************************************************** Enter a list of words one by one: a23456789 ^Z Only 1 string supplied. The length of string = 9

## N >= 2

*************************************************************** *** This program reports the longest and shortest strings *** *************************************************************** Enter a list of words one by one: a23456789 a23456 a2345b a234 a234567890 ^Z Vector size (number of words entered): 5 Length of shortest word: 4 Length of longest word: 10

Hi! I did this with a lot less code just by sorting the vector and accessing its first and last elements with random acccess, such as:

`cout << "the longest word is '" << list[0] << "' and its length is " << list[0].size() << " chars";`

Any particular reason why you didn’t go the sorting way?