Given an integer array nums and an integer val, remove all occurrences of val in nums in-place. The order of the elements may be changed. Then return the number of elements in nums which are not equal to val.Consider the number of elements in nums which are not equal to val be k, to get accepted, you need to do the following things:Change the array nums such that the first k elements of nums contain the elements which are not equal to val. The remaining elements of nums are not important as well as the size of nums.Return k.Custom Judge:The judge will test your solution with the following code:int[] nums = [...]; // Input arrayint val = ...; // Value to removeint[] expectedNums = [...]; // The expected answer with correct length. // It is sorted with no values equaling val.int k = removeElement(nums, val); // Calls your implementationassert k == expectedNums.length;sort(nums, 0, k); // Sort the first k elements of numsfor (int i = 0; i < actualLength; i++) { assert nums[i] == expectedNums[i];}If all assertions pass, then your solution will be accepted. Example 1:Input: nums = [3,2,2,3], val = 3Output: 2, nums = [2,2,_,_]Explanation: Your function should return k = 2, with the first two elements of nums being 2.It does not matter what you leave beyond the returned k (hence they are underscores).Example 2:Input: nums = [0,1,2,2,3,0,4,2], val = 2Output: 5, nums = [0,1,4,0,3,_,_,_]Explanation: Your function should return k = 5, with the first five elements of nums containing 0, 0, 1, 3, and 4.Note that the five elements can be returned in any order.It does not matter what you leave beyond the returned k (hence they are underscores). Constraints:0 <= nums.length <= 1000 <= nums[i] <= 500 <= val <= 100

Question

Given an integer array nums and an integer val, remove all occurrences of val in nums in-place. The order of the elements may be changed. Then return the number of elements in nums which are not equal to val.Consider the number of elements in nums which are not equal to val be k, to get accepted, you need to do the following things:Change the array nums such that the first k elements of nums contain the elements which are not equal to val. The remaining elements of nums are not important as well as the size of nums.Return k.Custom Judge:The judge will test your solution with the following code:int[] nums = [...]; // Input arrayint val = ...; // Value to removeint[] expectedNums = [...]; // The expected answer with correct length. // It is sorted with no values equaling val.int k = removeElement(nums, val); // Calls your implementationassert k == expectedNums.length;sort(nums, 0, k); // Sort the first k elements of numsfor (int i = 0; i < actualLength; i++) { assert nums[i] == expectedNums[i];}If all assertions pass, then your solution will be accepted. Example 1:Input: nums = [3,2,2,3], val = 3Output: 2, nums = [2,2,,]Explanation: Your function should return k = 2, with the first two elements of nums being 2.It does not matter what you leave beyond the returned k (hence they are underscores).Example 2:Input: nums = [0,1,2,2,3,0,4,2], val = 2Output: 5, nums = [0,1,4,0,3,,,_]Explanation: Your function should return k = 5, with the first five elements of nums containing 0, 0, 1, 3, and 4.Note that the five elements can be returned in any order.It does not matter what you leave beyond the returned k (hence they are underscores). Constraints:0 <= nums.length <= 1000 <= nums[i] <= 500 <= val <= 100

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Solution 1

This problem can be solved using a two-pointer technique. Here is a Python solution:

def removeElement(nums, val):
    i = 0
    for j in range(len(nums)):
        if nums[j] != val:
            nums[i] = nums[j]
            i += 1
    return i

This function works by initializing a pointer i at the start of the array. It then iterates over the array with another pointer j. If the element at j is not equal to val, it assigns the value at j to the position at i and then increments i. This effectively "removes" all instances of val from the array by overwriting them. The function then returns i, which is the number of elements not equal to val.

For example, if nums = [3,2,2,3] and val = 3, the function will return 2 and nums will be [2,2,_,_]. The underscores represent elements beyond the returned k, which are not important.

This problem has been solved

Solution 2

This problem can be solved using a two-pointer technique. Here is a Python solution:

def removeElement(nums, val):
    i = 0
    for j in range(len(nums)):
        if nums[j] != val:
            nums[i] = nums[j]
            i += 1
    return i

This function works by initializing a pointer i at the start of the array. It then iterates over the array with another pointer j. If the element at j is not equal to val, it assigns the value at j to the position i and then increments i. This effectively "shifts" all non-val elements to the front of the array. The function then returns i, which is the number of elements not equal to val.

For example, if nums = [3,2,2,3] and val = 3, the function will return 2 and nums will be modified to [2,2,_,_]. The underscores represent elements beyond the returned k, which are not important.

This problem has been solved

Solution 3

This problem is asking to remove all occurrences of a specific value from an array in-place and return the new length of the array. The order of the elements in the array can be changed.

Here is a step-by-step solution in Python:

Initialize a variable i to 0. This will be used to keep track of the position in the array where we can replace the current element if it is equal to val.

i = 0

Iterate over the array using a for loop.

for j in range(len(nums)):

Inside the loop, check if the current element is not equal to val. If it is not, replace the element at index i with the current element and increment i by 1.

    if nums[j] != val:
        nums[i] = nums[j]
        i += 1

After the loop, i will be the new length of the array. Return i.

return i

Here is the complete code:

def removeElement(nums, val):
    i = 0
    for j in range(len(nums)):
        if nums[j] != val:
            nums[i] = nums[j]
            i += 1
    return i

This function works by moving all elements not equal to val to the front of the array. The order of the elements is not preserved, but this is allowed by the problem statement. The space complexity is O(1) because no extra space is used, and the time complexity is O(n), where n is the length of the array, because each element is visited once.

This problem has been solved

Solution 4

Here is a Python solution for the problem:

def removeElement(nums, val):
    i = 0
    for j in range(len(nums)):
        if nums[j] != val:
            nums[i] = nums[j]
            i += 1
    return i

This function works by iterating over the list of numbers. If the current number is not equal to the value to be removed, it is moved to the next position in the "new" list (which is actually just the start of the original list). The variable i keeps track of the length of the "new" list.

For example, if nums = [3,2,2,3] and val = 3, the function will return 2 and nums will be modified to [2,2,2,3]. The first two elements are the ones not equal to val, and the rest of the list doesn't matter.

Similarly, if nums = [0,1,2,2,3,0,4,2] and val = 2, the function will return 5 and nums will be modified to [0,1,3,0,4,2,2,2]. The first five elements are the ones not equal to val, and the rest of the list doesn't matter.

This problem has been solved

Similar Questions

Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place such that each unique element appears only once. The relative order of the elements should be kept the same. Then return the number of unique elements in nums.Consider the number of unique elements of nums to be k, to get accepted, you need to do the following things:Change the array nums such that the first k elements of nums contain the unique elements in the order they were present in nums initially. The remaining elements of nums are not important as well as the size of nums.Return k.Custom Judge:The judge will test your solution with the following code:int[] nums = [...]; // Input arrayint[] expectedNums = [...]; // The expected answer with correct lengthint k = removeDuplicates(nums); // Calls your implementationassert k == expectedNums.length;for (int i = 0; i < k; i++) { assert nums[i] == expectedNums[i];}If all assertions pass, then your solution will be accepted. Example 1:Input: nums = [1,1,2]Output: 2, nums = [1,2,_]Explanation: Your function should return k = 2, with the first two elements of nums being 1 and 2 respectively.It does not matter what you leave beyond the returned k (hence they are underscores).Example 2:Input: nums = [0,0,1,1,1,2,2,3,3,4]Output: 5, nums = [0,1,2,3,4,_,_,_,_,_]Explanation: Your function should return k = 5, with the first five elements of nums being 0, 1, 2, 3, and 4 respectively.It does not matter what you leave beyond the returned k (hence they are underscores). Constraints:1 <= nums.length <= 3 * 104-100 <= nums[i] <= 100nums is sorted in non-decreasing order.

Given an integer array nums sorted in non-decreasing order, remove some duplicates in-place such that each unique element appears at most twice. The relative order of the elements should be kept the same.Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the first part of the array nums. More formally, if there are k elements after removing the duplicates, then the first k elements of nums should hold the final result. It does not matter what you leave beyond the first k elements.Return k after placing the final result in the first k slots of nums.Do not allocate extra space for another array. You must do this by modifying the input array in-place with O(1) extra memory.

You are given an integer array nums of size n and a positive integer k.Divide the array into one or more arrays of size 3 satisfying the following conditions:Each element of nums should be in exactly one array.The difference between any two elements in one array is less than or equal to k.Return a 2D array containing all the arrays. If it is impossible to satisfy the conditions, return an empty array. And if there are multiple answers, return any of them.

You are playing a game with your friend Ajay where Ajay gave you an array 'A' consisting of 'N' integers such that all the elements of the array does not exceed the value 'K'.Ajay has asked you to replace the minimum number of elements (replacement is the following operation: choose some index i from 1 to n and replace a[i] with some integer in range [1;k] ) to satisfy the following conditions:after all replacements, all a[i] are positive integers not greater than k;for all i from 1 to n/2 the following equation is true: a[i]+a[n]−i+1=x, where x should be the same for all n/2 pairs of elements.Note: It is guaranteed that 'N' is even.Input FormatThe first line contains two integers n and k — the length of a and the maximum possible value of some a[i] correspondingly. It is guratanteed that n is even .The second line of the test case contains n integers a1,a2,…,an, where ai is the i-th element of a.Constraints2≤n≤2⋅10^51≤k≤2⋅10^51≤a[i]≤kOutput FormatPrint the integer — the minimum number of elements you have to replace in a to satisfy the conditions from the problem statement.Sample Input 04 41 1 3 1Sample Output 01Sample Input 16 71 1 1 1 2 2Sample Output 11

You are given an array of positive integers nums.You need to select a subset of nums which satisfies the following condition:You can place the selected elements in a 0-indexed array such that it follows the pattern: [x, x2, x4, ..., xk/2, xk, xk/2, ..., x4, x2, x] (Note that k can be be any non-negative power of 2). For example, [2, 4, 16, 4, 2] and [3, 9, 3] follow the pattern while [2, 4, 8, 4, 2] does not.Return the maximum number of elements in a subset that satisfies these conditions.

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