Is there any built in function to check whether given two numbers are in the same order in a given integer array?

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I want to check whether given two integers in a specific order are in the same order in a given integer array. I wonder whether there is an easy way to do this like a built-in CPP method. If there is no built-in method, suggest me an efficient way to do this as I have a few sets of two integers (not only one set) to check over one array.

given two numbers: 8 3 given array: 2 8 6 1 3 9 output: YES

You could do something like

bool check(std::pair<int, int> numbers = {8, 3},
std::array<int, 6> arr = {2, 8, 6, 1, 3, 9}) {
    if (numbers.first != numbers.second)
        return std::find(std::find(std::begin(arr), std::end(arr), numbers.first), std::end(arr), numbers.second) == std::end(arr);
    return std::count(std::begin(arr), std::end(arr), numbers.first) >= 2;
}

If both numbers are different the inner find searches for the first value. The outer find starts at the position of the first value and searches for the second value.

Else the count is checked.

Check if all values of array are equal, How do you check if an array has the same value? Given an array A[] and a number x, check for pair in A[] with sum as x Write a program that, given an array A[] of n numbers and another number x, determines whether or not there exist two elements in S whose sum is exactly x.

You could also try:

    std::array<int, 6> content = {2, 3, 6, 1, 8, 9};
    auto lookup = [content](int a, int b)
    {
        return std::distance(std::find(content.begin(), content.end(), a), std::find(content.rbegin(), content.rend(), b));
    };
    lookup(8, 3);

lookup will be positive if 8 comes before 3 and negative otherwise.

How to Find all Pairs in Array of Integers Whose sum is Equal to a , How do you find all pairs of an integer array whose sum is equal to a given number? Python xrange() to check integer in between two numbers This method (xrange()) would only work in Python 2.7 or below. But since Python 2.7 is still in use, so we are giving an example of the same.

Search the entire container to find the first one. Search from the position of the first one to the end of the container to find the second one. If that search succeeds, they're in the expected order. If not, they're not.

int first_value = 8;
int second_value = 3;
std::array<int, 6> values = { 2, 8, 6, 1, 3, 9 };

auto first_pos = std::find(values.begin(), values.end(), first_value);
if (first_pos != values.end())
    ++first_pos;
auto second_pos = std::find(first_pos, values.end(), second_value);
if (second_pos != values.end())
    std::cout << "YES\n";

Program to check if an array is sorted or not (Iterative and Recursive , How do you check if an array is already sorted? Given two integer number n and d. The task is to find the number between 0 to n which contain the specific digit d. Examples: Input : n = 20 d = 5 Output : 5 15 Input

Use adjacent find. I suppose find_if could also do the job.

Find the two repeating elements in a given array, , n will eventually fall to one, satisfying Step 1. Given two arrays: arr1[0..m-1] and arr2[0..n-1]. Find whether arr2[] is a subset of arr1[] or not. Both the arrays are not in sorted order. It may be assumed that elements in both array are distinct.

Given an array A[] and a number x, check for pair in A[] with sum as x , How do you find an element is present in an array or not? Simple Approach: The idea is to run a loop from start to end and for every index in the given string check whether the sub-string can be formed from that index. This can be done by running a nested loop traversing the given string and in that loop run another loop checking for sub-string from every index.

Computational Number Theory: Proceedings of the Colloquium on , And all elements occur once except two numbers which occur twice. when you see an element whose count is already set, print it as duplicate. This method uses the range given in the question to restrict the size of count[], but doesn't use the data that We know the sum of integers from 1 to n is n(n+1)/2 and product is n! Program to count digits in an integer (4 Different Methods) Count the number of digits in a long integer entered by a user. Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++ Made Easy, Write a program that, given an array A[] of n numbers and another number x, determines Duration: 10:35 Posted: 10 Oct 2011 In the above program, we've used a non-primitive data type String and used Arrays's stream() method to first convert it to a stream and anyMatch() to check if array contains the given value toFind. Share on:

Comments
  • Find first of the first number and find last of the second number. Check if first index is smaller than second.
  • You can use two std::find calls to do this. For more information on how to use the C++ library's algorithms, including std::find, you should find plenty of information and examples in your C++ book.
  • Can the array contain duplicates?
  • No standard CPP function comes to mind, but there may be some combination of them that works.
  • That's what I described in the comments but I didn't know how to implement it. I didn't know that normal iterators and reverse iterators could be compared like this.
  • I had the same idea but it gives a wrong result for first_value = 8; second_value = 8;
  • How would you use adjacent_find to do this?
  • Oops my mistake, he is not looking for consecutive elements. Only find_if or handwritten algorithm then.