Count of indices with value 1 after performing given operations sequentially

Our objective is to successfully confront the presented issue by determining the number of indices with a value of 1 following consecutive operations. We have planned to accomplish this task through sequential and methodical execution of each operation utilizing C++ as our preferred programming language.

However, for a permanent solution to be achieved, it is crucial for us to formulate an effective algorithmic blueprint coded appropriately.

Syntax

To better prepare ourselves for the algorithm. It is advisable to become acquainted with the coding syntax. The code snippets that follow will make use of a particular method as shown below −

// Syntax for the method used in the following codes
int countIndicesWithOne(const vector& nums) {
   // Implementation of the method
}

Algorithm

To solve the problem, we will follow a step-by-step algorithm. The algorithmic approach will help us systematically perform the required operations and count the indices with a value of 1. The following steps outline the algorithm −

• Initialize a variable, let's call it count, to keep track of the count of indices with a value of 1.

• Iterate through the given array nums.

• For each element in nums, perform the following −

• If the element is equal to 1, increment the count variable by 1.

• Return the value of count after iterating through all elements.

• With a defined algorithm at our disposal, let us now review two distinct methods to proficiently tackle the issue at hand.

Approaches

Now, let's discuss two different approaches −

Approach 1: Linear Scan

In order to count how many indices within our array contain a value of 1, we will utilize an elementary linear scanning process. Through traversing all indexes systematically, and pinpointing those containing simply one element with a stored '1', an accurate final count can be tabulated. The segment provided below outlines part of this solution −

Example

#include <iostream>
#include <vector>

using namespace std;

int countIndicesWithOne(const vector<int>& nums) {
   int count = 0;
   for (int num : nums) {
      if (num == 1) {
         count++;
      }
   }
   return count;
}

int main() {
   vector<int> nums = {1, 0, 1, 1, 0, 1};
   int result = countIndicesWithOne(nums);
   cout << "Count of indices with value 1: " << result << endl;
   return 0;
}

Output

Count of indices with value 1: 4

Explanation

The initial approach for counting indices with a value of 1 in an array calls for employing a linear scan method. The process begins by initiating 'count', which functions as a counter variable throughout the code. Next, an iteration over every element contained within the array commences via a for loop construct. For each element encountered during this exploration phase, we test if its value is equal to one before increasing count accordingly whenever one such occurrence presents itself. After examining all individual elements within the array declared initially within our approach number one, we sum up every instance that satisfies our criterion into 'count,' which finally ends up returned at completion while providing an uncomplicated yet sound methodology for approaching such problems while reducing complexity; ideal when trying out development ideas.

Approach 2: Bit Manipulation

An efficient way to compute the prevalence rate of 1s in indices is by utilizing various bit manipulation methods, which helps us keep track quickly and systematically. Using this specific technique proves exceptionally helpful when managing multiple integers simultaneously. Check out the following sample code designed for demonstrating how we put it all together using our innovative methodology −

Example

#include <iostream>
#include <vector>

using namespace std;

int countIndicesWithOne(const vector<int>& nums) {
   int count = 0;
   for (int num : nums) {
      count += __builtin_popcount(num);
   }
   return count;
}

int main() {
   vector<int> nums = {1, 0, 1, 1, 0, 1};
   int result = countIndicesWithOne(nums);
   cout << "Count of indices with value 1: " << result << endl;
   return 0;
}

Output

Count of indices with value 1: 4

Explanation

Our second approach uses bit manipulation techniques to count how many indices have a value equal to one in an inputted array successfully. This is done by initializing a variable called `count`, whose purpose is to keep track of how many values are counted as having a value equal to one. Similar to our first approach, we iterate through all elements within our array using a for-loop. We apply an efficient method using "__builtin_popcount" instead of direct conditioning check "if (value==1)", by counting only set bits or ones in binary representation under that element and add it up iteratively into our initialized variable used earlier "count." Once all elements are checked within our current loop/, we return back its final index 'one' valued positions as represented by 'counts'. In summary ,we can now conclude that this method provides us with better optimized solutions while dealing with arrays since it utilizes bitwise operations as well.

Conclusion

We conducted a comprehensive investigation using established procedures to address the daunting task of calculating the quantity of C++ indices that have a value of 1. We proficiently presented two highly-effective methodologies - linear scan and bit manipulation - complete with functional code for each strategy that can readily be integrated into your projects. Our approach entailed defining the required syntax and algorithm to lay out these methods. By utilizing these techniques, you will be able to tally the requisite indices with keen accuracy and extend this knowledge in multiple contexts.