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To solve the problem of finding the missing element in an array A consisting of N different integers from the range [1..(N + 1)], you can utilize the mathematical approach of calculating the expected sum versus the actual sum of the array elements.

Here’s the step-by-step explanation of the approach:

The sum of the first N+1 integers (which are [1, 2, ..., N+1]) can be calculated using the formula:

This formula gives the sum of all integers from 1 to N+1.

Compute the sum of all elements in array A.

The missing element will be the difference between the expected sum and the actual sum of array A.

Given the constraints (N can be up to 100,000), this approach is efficient with a time complexity of O(N) because it involves a single pass through the array to compute the sum.

Here is the implementation of the solution in JavaScript:

function solution(A) {
    const N = A.length;
    const expectedSum = (N + 1) * (N + 2) / 2;
    let actualSum = 0;
    
    for (let i = 0; i < N; i++) {
        actualSum += A[i];
    }
    
    const missingElement = expectedSum - actualSum;
    return missingElement;
}

// Example usage:
const A = [2, 3, 1, 5];
console.log(solution(A)); // Output: 4

Explanation of the Example:

For the array A = [2, 3, 1, 5], where N = 4 (since A.length = 4), the expected sum of [1, 2, 3, 4, 5] is 15.

The actual sum of elements in A is 2 + 3 + 1 + 5 = 11.

Therefore, the missing element is 15 - 11 = 4.

This approach ensures that you find the missing element in linear time, making it optimal for large inputs as specified in the problem constraints.

To solve the problem of determining the minimal number of jumps a frog needs to make from position X to reach or exceed position Y with each jump of distance D, we can break down the solution as follows:

Understanding the Problem:

• Inputs:

  • X: Starting position of the frog.
  • Y: Target position the frog wants to reach or exceed.
  • D: Fixed distance that the frog can jump in one move.

• Output:

  • Number of jumps required for the frog to reach or exceed Y starting from X.

If distance is exactly divisible by D (i.e., distance % D == 0), then the number of jumps required is:

If distance is not exactly divisible by D, then the frog needs one additional jump to cover the remaining distance:

Approach:

We determine how far the frog needs to jump to reach Y from X. This can be found using:

Then we calculate the number of jumps required to cover this distance:

Here, (\left\lceil x \right\rceil) denotes the ceiling function, which rounds up to the nearest integer.

Implementation in JavaScript:

Here’s how you can implement this in JavaScript:

function solution(X, Y, D) {
    const distance = Y - X;
    if (distance % D === 0) {
        return distance / D;
    } else {
        return Math.ceil(distance / D);
    }
}

We compute distance as Y - X.

Use modulo operation (%) to check if distance is exactly divisible by D.

If divisible, return distance / D.

If not divisible, use Math.ceil(distance / D) to round up to the nearest integer.

Example Usage

console.log(solution(10, 85, 30)); // Output: 3

• For X = 10, Y = 85, and D = 30, the frog needs 3 jumps:
  • 1st jump: 10 + 30 = 40
  • 2nd jump: 40 + 30 = 70
  • 3rd jump: 70 + 30 = 100 (exceeds 85, so it’s sufficient).

This solution efficiently computes the minimal number of jumps required using basic arithmetic operations, ensuring optimal performance even for large inputs within the specified constraints.

To solve this problem efficiently, we can utilize the properties of the XOR bitwise operation. XOR has a useful property where:

This means that when you XOR all elements of an array together, elements that appear an even number of times will cancel out to zero, leaving only the element that appears an odd number of times.

First, start with a variable initialized to zero. This will be used to XOR all elements of the array.

Next, iterate through each element of the array and XOR it with the variable initialized in the previous step.

After iterating through the array, the variable will contain the value of the unpaired element because all paired elements will cancel each other out due to their even occurrences.

Let’s see this approach implemented in JavaScript:

function solution(A) {
    let result = 0;
    
    for (let i = 0; i < A.length; i++) {
        result ^= A[i];
    }
    
    return result;
}

Explanation of the Code:

result is initialized to 0. This will be our accumulator for the XOR operation.

We loop through each element of the array A.

We XOR (^=) each element with result. This effectively accumulates all XOR operations across the array.

• Return the result: After iterating through the array, result will contain the value of the unpaired element because all paired elements cancel out.

Efficiency:

This algorithm operates in ( O(N) ) time complexity, where ( N ) is the number of elements in the array. This is optimal for the problem constraints (with ( N ) up to 1,000,000), ensuring that the solution is both time-efficient and space-efficient.

Edge Cases:

The algorithm handles arrays of minimum size (e.g., single element) and ensures correct results for all valid inputs within the specified constraints.

This method leverages XOR’s properties to solve the problem in a straightforward and efficient manner.

To solve the problem of rotating an array A by K positions to the right, we can follow a straightforward approach:

Steps to Approach

If A is empty (N = 0), return an empty array since there’s nothing to rotate.

If K is zero, return A as it is since no rotation is needed.

Since rotating an array N times where N is the length of the array results in the same array, the effective number of rotations K can be reduced to K % N. This is because rotating an array N times or any multiple of N results in the array itself.

To rotate the array to the right by K positions, the idea is to move the last K elements of A to the front and shift the rest to the right.

• This can be achieved by creating a new array where the elements from A[N-K] to A[N-1] (last K elements) are placed at the beginning, followed by the elements from A[0] to A[N-K-1] (remaining elements).

Implementation in JavaScript

Here’s the implementation based on the outlined approach:

function solution(A, K) {
    const N = A.length;
    
    // Handle edge cases
    if (N === 0 || K === 0) {
        return A;
    }
    
    // Effective rotations
    K = K % N;
    
    // If no effective rotation needed
    if (K === 0) {
        return A;
    }
    
    // Create a new array for rotated result
    const rotatedArray = [];
    
    // Elements from A[N-K] to A[N-1] (last K elements)
    for (let i = N - K; i < N; i++) {
        rotatedArray.push(A[i]);
    }
    
    // Elements from A[0] to A[N-K-1] (first N-K elements)
    for (let i = 0; i < N - K; i++) {
        rotatedArray.push(A[i]);
    }
    
    return rotatedArray;
}

Handle cases where A is empty or K is zero directly to optimize and prevent unnecessary calculations.

By using K % N, we ensure that we only rotate as many times as necessary (K rotations effectively reduces to rotating K % N times).

We split the array into two parts.

Elements from A[N-K] to A[N-1] are appended first (last K elements).

Elements from A[0] to A[N-K-1] are appended next (remaining elements).

This approach effectively constructs the rotated array in a linear pass through the original array A.

This implementation ensures correctness by adhering to the problem requirements and handles all specified edge cases.

To solve the problem of finding the longest binary gap in a given positive integer ( N ), we can follow these steps:

1. Convert the integer ( N ) into its binary representation. In JavaScript, you can do this using toString(2) method.

2. Traverse through the binary representation to identify sequences of consecutive zeros (‘0’). Keep track of the length of each sequence.

3. As we identify these sequences, keep track of the maximum length encountered.

4. Finally, return the length of the longest binary gap found. If no binary gaps are found, return 0.

Here’s the JavaScript implementation based on the described approach:

function solution(N) {
    const binaryString = N.toString(2); // Convert N to binary string
    let maxGapLength = 0;
    let currentGapLength = 0;
    let inGap = false;
    
    for (let i = 0; i < binaryString.length; i++) {
        if (binaryString[i] === '1') {
            if (inGap) {
                // We've encountered the end of a gap
                if (currentGapLength > maxGapLength) {
                    maxGapLength = currentGapLength;
                }
                currentGapLength = 0;
            }
            inGap = true;
        } else {
            if (inGap) {
                // We are inside a gap
                currentGapLength++;
            }
        }
    }
    
    return maxGapLength;
}

N.toString(2) converts ( N ) to its binary string representation.

Then we ierate through each character of the binary string

When encountering ‘1’, check if we were already in a gap (i.e., inGap is true). If so, this signifies the end of a gap, so update maxGapLength if the current gap is longer than previously recorded.

When encountering ‘0’, if we are in a gap (inGap is true), increment currentGapLength.

After the loop, ensure to check if the last recorded gap (if any) is the longest.

This algorithm is efficient with a time complexity of ( O(log N) ), where ( N ) is the integer input, due to the conversion to binary and subsequent traversal of its digits. This ensures it can handle the maximum input size within reasonable time constraints.