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Write the JAVA code for the below problem statement: You are given an integer array nums and an integer k. Append k unique positive integers that do not appear in nums to nums such that the resulting total sum is minimum. Return the sum of the k integers appended to nums. Example 1: Input: nums = [1,4,25,10,25], k = 2 Output: 5 Explanation: The two unique positive integers that do not appear in nums which we append are 2 and 3. The resulting sum of nums is 1 + 4 + 25 + 10 + 25 + 2 + 3 = 70, which is the minimum. The sum of the two integers appended is 2 + 3 = 5, so we return 5. Example 2: Input: nums = [5,6], k = 6 Output: 25 Explanation: The six unique positive integers that do not appear in nums which we append are 1, 2, 3, 4, 7, and 8. The resulting sum of nums is 5 + 6 + 1 + 2 + 3 + 4 + 7 + 8 = 36, which is the minimum. The sum of the six integers appended is 1 + 2 + 3 + 4 + 7 + 8 = 25, so we return 25. Constraints: 1 <= nums.length <= 10^5 1 <= nums[i] <= 10^9 1 <= k <= 10^8 Use the below template: class Solution { public long minimalKSum(int[] nums, int k) { } }

Here's the JAVA code to solve the problem statement:

java

import java.util.Arrays;

class Solution {
    public long minimalKSum(int[] nums, int k) {
        // Sort the given array
        Arrays.sort(nums);

        int n = nums.length;
        int i = 0;
        long sum = 0;

        // Loop through each of the k missing numbers and add them to the sum
        while (k > 0) {
            int nextNum = i + 1;
            if (nextNum < n && nums[i] == nums[nextNum]) {
                // If the current number is already present in the array, move to the next number
                i++;
                continue;
            }
            // Add the missing number to the sum and decrement k
            sum += nums[i] + nextNum;
            i++;
            k--;
        }

        return sum;
    }
}

Explanation:

  1. We start by sorting the given array of integers in ascending order.
  2. We initialize two variables i and sum. i is used to keep track of the current index of the array and sum is used to keep track of the sum of the k missing numbers.
  3. We loop through each of the k missing numbers and add them to the sum.
  4. For each missing number, we check if the next number in the sorted array is equal to the current number. If it is, we move to the next number in the array.
  5. If the next number is not equal to the current number, we add the missing number to the sum and decrement k.
  6. Finally, we return the sum of the k missing numbers.

Modify the above code such that the below Runtime error does not occur: Runtime Error java.lang.ArrayIndexOutOfBoundsException: Index 2 out of bounds for length 2 at line 19, Solution.minimalKSum at line 54, __DriverSolution__.__helper__ at line 87, __Driver__.main

The given runtime error occurs because the code tries to access an array element that is outside the bounds of the array. This can happen when the code tries to access an element with an index greater than or equal to the length of the array.

To fix this error, we need to make sure that the index i stays within the bounds of the array. We can modify the while loop condition to check if both i and i+1 are within the bounds of the array before comparing nums[i] and nums[i+1].

Here's the modified code that should fix the runtime error:

java

import java.util.Arrays;

class Solution {
    public long minimalKSum(int[] nums, int k) {
        // Sort the given array
        Arrays.sort(nums);

        int n = nums.length;
        int i = 0;
        long sum = 0;

        // Loop through each of the k missing numbers and add them to the sum
        while (k > 0) {
            int nextNum = i + 1;
            if (nextNum < n && (i == n-1 || nums[i] != nums[nextNum])) {
                // If the current number is already present in the array, move to the next number
                // or if i is at the end of the array, we can add the missing number to the sum
                sum += nums[i] + nextNum;
                k--;
            }
            i++;
        }

        return sum;
    }
}

Explanation:

  1. We start by sorting the given array of integers in ascending order.
  2. We initialize three variables i, n, and sum. i is used to keep track of the current index of the array, n is the length of the array, and sum is used to keep track of the sum of the k missing numbers.
  3. We loop through each of the k missing numbers and add them to the sum.
  4. For each missing number, we check if the next number in the sorted array is equal to the current number. If it is, we move to the next number in the array.
  5. If the next number is not equal to the current number or i is at the end of the array, we add the missing number to the sum and decrement k.
  6. Finally, we return the sum of the k missing numbers.

Modify the above code such that the below runtime error does not occur: Runtime Error java.lang.ArrayIndexOutOfBoundsException: Index 2147483647 out of bounds for length 2 at line 13, Solution.minimalKSum at line 54, __DriverSolution__.__helper__ at line 87, __Driver__.main

The given runtime error occurs because the code is trying to access an array element at index 2147483647, which is out of the bounds of the array. This can happen when the input value of k is greater than the number of missing positive integers that need to be appended to the array.

To fix this error, we can modify the loop condition to make sure that we stop looping once we have appended all the missing positive integers to the array, even if k is still greater than 0.

Here's the modified code that should fix the runtime error:

java

import java.util.Arrays;

class Solution {
    public long minimalKSum(int[] nums, int k) {
        // Sort the given array
        Arrays.sort(nums);

        int n = nums.length;
        int i = 0;
        long sum = 0;

        // Loop through each of the k missing numbers and add them to the sum
        while (k > 0 && i < n) {
            int nextNum = i + 1;
            if (nextNum < n && (i == n-1 || nums[i] != nums[nextNum])) {
                // If the current number is already present in the array, move to the next number
                // or if i is at the end of the array, we can add the missing number to the sum
                sum += nums[i] + nextNum;
                k--;
            }
            i++;
        }

        return sum;
    }
}

Explanation:

  1. We start by sorting the given array of integers in ascending order.
  2. We initialize three variables i, n, and sum. i is used to keep track of the current index of the array, n is the length of the array, and sum is used to keep track of the sum of the k missing numbers.
  3. We loop through each of the k missing numbers and add them to the sum, as long as i is less than n and k is greater than 0.
  4. For each missing number, we check if the next number in the sorted array is equal to the current number. If it is, we move to the next number in the array.
  5. If the next number is not equal to the current number or i is at the end of the array, we add the missing number to the sum and decrement k.
  6. Finally, we return the sum of the k missing numbers.

The following test cases failed: Input nums = [1,4,25,10,25] k = 2 Output 8 Expected 5 Input nums = [5,6] k = 6 Output 6 Expected 25 Also, use the below hints: 1. The k smallest numbers that do not appear in nums will result in the minimum sum. 2. Recall that the sum of the first n positive numbers is equal to n * (n+1) / 2. 3. Initialize the answer as the sum of 1 to k. Then, adjust the answer depending on the values in nums.

Thank you for bringing this to our attention. It seems that there is a mistake in the previous solution. Instead of adding the missing numbers to the sum, we need to add the sum of the first k positive integers to the sum, and then subtract the sum of the k smallest positive integers that are already present in the given array.

Here's the corrected code:

java

import java.util.Arrays;

class Solution {
    public long minimalKSum(int[] nums, int k) {
        // Sort the given array
        Arrays.sort(nums);

        int n = nums.length;
        int i = 0;
        long sum = (long) k * (k+1) / 2;

        // Subtract the sum of the k smallest positive integers in nums
        while (k > 0 && i < n) {
            if (nums[i] <= k) {
                sum -= nums[i];
                k--;
            } else {
                break;
            }
            i++;
        }

        return sum;
    }
}

Explanation:

  1. We start by sorting the given array of integers in ascending order.
  2. We initialize three variables i, n, and sum. i is used to keep track of the current index of the array, n is the length of the array, and sum is used to keep track of the sum of the k missing numbers.
  3. We calculate the sum of the first k positive integers using the formula n * (n+1) / 2 and assign it to the variable sum.
  4. We loop through the array and subtract the value of each number from the sum as long as the number is less than or equal to k. We also decrement k for each number that we subtract.
  5. Finally, we return the sum, which now represents the sum of the k missing positive integers.

With this implementation, the given test cases should pass.

Follow the below steps to write the JAVA code: -> first we sort the array -> then we add/append elements in array which are missing in the range 1 2 3 . . . till kth missing element -> then we check if nums[i+1]-nums[i] >1 , means ...missing elements are present . -> for finding sum we'll use formula of finding sum of an AP : ( (first term+last term) * total term )/2

Here's the Java code implementation based on the given steps:

java

import java.util.Arrays;

class Solution {
    public long minimalKSum(int[] nums, int k) {
        Arrays.sort(nums);
        int n = nums.length;
        int i = 0;
        long sum = 0;
        
        // Find the k missing elements
        for (int j = 1; j <= k; j++) {
            while (i < n && nums[i] == j) {
                i++;
            }
            if (i == n || nums[i] > j) {
                sum += j;
            } else {
                j--;
            }
        }
        
        // Add the remaining elements in the array
        while (i < n) {
            if (i == 0 || nums[i] > nums[i-1] + 1) {
                sum += nums[i];
            }
            i++;
        }
        
        // Calculate the sum of the k missing elements
        sum += ((long)k * (k+1) / 2);
        sum -= ((long)(k-i+1) * (k-i+2) / 2);
        
        return sum;
    }
}

Explanation:

  1. We start by sorting the given array of integers in ascending order using the Arrays.sort method.
  2. We initialize three variables i, n, and sum. i is used to keep track of the current index of the array, n is the length of the array, and sum is used to keep track of the sum of the missing numbers.
  3. We loop through the integers from 1 to k and check if they are present in the given array. If they are not present, we add them to the sum. If they are present, we decrement j by 1 and continue.
  4. We then loop through the remaining elements in the array and add them to the sum if they are not already present in the sum. We use the nums[i-1] + 1 condition to check if there are any missing elements between the current element and the previous element.
  5. We then calculate the sum of the missing elements using the formula for the sum of an arithmetic progression.
  6. Finally, we subtract the sum of the elements already present in the array from the sum of the missing elements to get the sum of the k appended elements.

With this implementation, the given test cases should pass.

Write the JAVA program for this problem statement by simply using the formula n*(n+1)/2 and use Set data structure.

Sure, here's the updated code that uses the formula n*(n+1)/2 and Set data structure:

python

import java.util.*;

class Solution {
    public int minimalKSum(int[] nums, int k) {
        Set<Integer> set = new HashSet<>();
        for (int num : nums) {
            set.add(num);
        }
        int sum = k * (k + 1) / 2;
        for (int i = 1; i <= k; i++) {
            if (set.contains(i)) {
                sum -= i;
                k++;
            }
        }
        return sum;
    }
}

Explanation:

  1. We first create a Set to store all the numbers in the nums array.
  2. We calculate the sum of the first k positive integers using the formula k*(k+1)/2.
  3. We loop through the integers from 1 to k, and check if each integer is present in the Set.
  4. If the integer is present in the Set, we subtract it from the sum and increment k by 1, since we need to append another integer to make up for the missing one.
  5. Finally, we return the sum, which is the sum of the k integers that were appended to the nums array.

Modify the above code and use the below trick: We sort numbers, and then swipe the range from 1 to 10^9, appending numbers that do not appear in the array. However, this will cause TLE, since k can be very large. Compute the minimal possible sum as k * (k + 1) / 2). Then, we go through unique numbers in the array, and substitute all numbers less than, or equal k, with increasing numbers greater than k..

Sure, here's the modified code that uses the given trick:

csharp

import java.util.*;

class Solution {
    public int minimalKSum(int[] nums, int k) {
        Arrays.sort(nums);
        Set<Integer> set = new HashSet<>();
        for (int num : nums) {
            set.add(num);
        }
        int sum = k * (k + 1) / 2;
        for (int i = 0; i < nums.length && k > 0; i++) {
            if (nums[i] <= k && !set.contains(k)) {
                set.remove(nums[i]);
                set.add(k);
                sum += k - nums[i];
                k--;
            }
        }
        return sum;
    }
}

Explanation:

  1. We first sort the nums array in ascending order.
  2. We create a Set to store all the numbers in the nums array.
  3. We calculate the sum of the first k positive integers using the formula k*(k+1)/2.
  4. We loop through the nums array, and if we encounter a number less than or equal to k that is not present in the Set, we replace it with k, and update the sum accordingly.
  5. We decrement k by 1, since we have now appended one of the missing integers to the nums array.
  6. Finally, we return the sum, which is the sum of the k integers that were appended to the nums array.

Write the JAVA code by using the below steps: 1. Sort the array in ascending numerical order; 2. Use all values ​​that occur up to the minimum value in the array; 3. Check and use values ​​that are in the range of the array (from the minimum value of the array to the maximum value of the array) and are not used in it; 4. Use all values ​​occurring after the maximum value in the array; 5. At each step, compare the value of the counter and "k".

Here's the Java code implementing the steps mentioned above:

arduino

import java.util.*;

class Solution {
    public long minimalKSum(int[] nums, int k) {
        Arrays.sort(nums);
        Set<Integer> used = new HashSet<>();
        int min = nums[0];
        long sum = 0;
        
        for (int i = 0; i < nums.length && k > 0; i++) {
            if (nums[i] <= min + k - 1) {
                used.add(nums[i]);
            } else if (nums[i] > min + k - 1 && nums[i] <= min + 2 * k - 1) {
                if (!used.contains(nums[i])) {
                    used.add(nums[i]);
                    k--;
                }
            }
            sum += nums[i];
        }
        
        int max = nums[nums.length - 1];
        for (int i = 1; i <= k; i++) {
            int num = max + i;
            while (used.contains(num)) {
                num++;
            }
            used.add(num);
            sum += num;
        }
        
        return sum - (long)k * (k + 1) / 2;
    }
}

The following test case got falled: Input nums = [1,4,25,10,25] k = 2 Output 90 Expected 5 Input nums = [5,6] k = 6 Output 47 Expected 25

Here is the modified code which should work for the given test cases:

csharp

import java.util.*;

class Solution {
    public long minimalKSum(int[] nums, int k) {
        Arrays.sort(nums);  // Sort the given array
        int n = nums.length;
        Set<Integer> set = new HashSet<>();  // Use a Set to keep track of unique numbers in the array
        
        // Add all elements of the array to the Set
        for (int num : nums) {
            set.add(num);
        }
        
        long sum = 0;
        
        // Add all numbers up to the minimum value in the array
        for (int i = 1; i < nums[0]; i++) {
            set.add(i);
            sum += i;
            k--;
            if (k == 0) {
                return sum;
            }
        }
        
        // Check and use values that are in the range of the array and are not used in it
        for (int i = 1; i < n; i++) {
            int diff = nums[i] - nums[i-1];  // Calculate the difference between consecutive elements
            if (diff > 1) {
                for (int j = 1; j < diff && k > 0; j++) {
                    int num = nums[i-1] + j;
                    if (!set.contains(num)) {
                        set.add(num);
                        sum += num;
                        k--;
                        if (k == 0) {
                            return sum;
                        }
                    }
                }
            }
        }
        
        // Add all numbers after the maximum value in the array
        for (int i = nums[n-1]+1; k > 0; i++) {
            if (!set.contains(i)) {
                set.add(i);
                sum += i;
                k--;
            }
        }
        
        return sum;
    }
}

The memory limit exceeded for the above code for a very large input. Rectify it.

To reduce the memory usage, instead of using a HashSet to store all the values that appear in the array, we can use a boolean array to mark the values that appear. The boolean array will have a size of 10^9, which is the maximum possible value in the input. This should significantly reduce the memory usage.

Here is the modified code:

arduino

class Solution {
    public long minimalKSum(int[] nums, int k) {
        Arrays.sort(nums);
        boolean[] appeared = new boolean[1000000001];
        for (int num : nums) {
            appeared[num] = true;
        }
        int i = 1;
        long sum = 0;
        while (k > 0) {
            if (!appeared[i]) {
                sum += i;
                k--;
            }
            i++;
        }
        return sum;
    }
}

This should fix the memory limit issue.

The memory limit exceeded for the above code.

To optimize the memory usage, we can replace the HashSet with a boolean array to keep track of the numbers that have been used. This will use much less memory than a hash set, especially when the range of the numbers is very large. Here's the modified code:

css

class Solution {
    public long minimalKSum(int[] nums, int k) {
        Arrays.sort(nums);
        int n = nums.length;
        boolean[] used = new boolean[(int)1e9+1];
        for (int i = 0; i < n && nums[i] <= k; i++) {
            used[nums[i]] = true;
        }
        long sum = k * (long)(k+1) / 2;
        for (int i = 0; i < n && nums[i] <= k; i++) {
            if (!used[nums[i]]) {
                sum += nums[i];
                k--;
                used[nums[i]] = true;
            }
        }
        for (int i = 1; i <= (int)1e9 && k > 0; i++) {
            if (!used[i]) {
                sum += i;
                k--;
            }
        }
        return sum;
    }
}

Note that we create a boolean array used with length (int)1e9+1 to keep track of the numbers that have been used. We also use the long data type for the sum variable to avoid integer overflow.

Use the below sample code as a hint to write the JAVA code: Arrays.sort(nums); Set set = new HashSet<>(); long sum = 0; for (int num: nums) { if (!set.contains(num) && num <= k) { k++; sum += num; } set.add(num); } long res = (long)(1 + k) * k / 2 - sum; return res;

Here's the updated JAVA code using the above hint:

arduino

import java.util.*;

public class Solution {
    public static long minimalKSum(int[] nums, int k) {
        Arrays.sort(nums);
        Set<Integer> set = new HashSet<>();
        long sum = 0;

        for (int num: nums) {
            if (!set.contains(num) && num <= k) {
                k++;
                sum += num;
            }
            set.add(num);
        }

        long res = (long)(1 + k) * k / 2 - sum;
        return res;
    }
}

This code uses a set to keep track of the unique elements in the array, and adds missing elements to the sum. Finally, it calculates the minimal possible sum as the sum of all numbers up to k minus the sum of elements in the array.