CareerCup

void editArray(int arr[], int m, int arr_size)
{
	int cur, i, cur_ele = -1;
	int count = 0;
    
	cur = 0;
	
	for(int i = 0;i < arr_size;i++)
	{
		if (cur_ele != arr[i])
		{
			cur_ele = arr[i];
			count = 0;
		}

		if (cur_ele == arr[i] && count < min(m, 2))
		{
            		count++;
            		arr[cur] = arr[i];
            		cur++;
        	}
    	}
    
    	for(int i = cur;i < arr_size;i++)
        	arr[i] = 0;
}

can you add an example?

if an array is {2,2,2,2,3} and the method is editArray(arr[],m)
if m=4
then after the method array should be {2,2,3,0,0}

/**
	 * 
	 * @param input Array of positive numbers in the sorted order.
	 * @param x 
	 */
	private static void editArray(int[] input, int x) {

		int lCount = 0;
		for (int i = 0, j = -1, k = 0; i < input.length; i++) {

			if (input[i] == x) {

				lCount++;
				if (lCount > 2) {
					if (j == -1) {
						j = i;
					}
					input[i] = 0;
				}
			} else if (lCount > 0) {

				if (j != -1) {
					k = j;
					j = -1;
				}
				input[k++] = input[i];
				input[i] = 0;
			}

		}
	}

/**
	 * 
	 * @param input Array of positive numbers in the sorted order.
	 * @param x 
	 */
	private static void editArray(int[] input, int x) {

		int lCount = 0;
		for (int i = 0, j = -1, k = 0; i < input.length; i++) {

			if (input[i] == x) {

				lCount++;
				if (lCount > 2) {
					if (j == -1) {
						j = i;
					}
					input[i] = 0;
				}
			} else if (lCount > 0) {

				if (j != -1) {
					k = j;
					j = -1;
				}
				input[k++] = input[i];
				input[i] = 0;
			}

		}
	}

/**
	 * 
	 * @param input Array of positive numbers in the sorted order.
	 * @param x 
	 */
	private static void editArray(int[] input, int x) {

		int lCount = 0;
		for (int i = 0, j = -1, k = 0; i < input.length; i++) {

			if (input[i] == x) {

				lCount++;
				if (lCount > 2) {
					if (j == -1) {
						j = i;
					}
					input[i] = 0;
				}
			} else if (lCount > 0) {

				if (j != -1) {
					k = j;
					j = -1;
				}
				input[k++] = input[i];
				input[i] = 0;
			}

		}
	}

/**
	 * 
	 * @param input Array of positive numbers in the sorted order.
	 * @param x 
	 */
	private static void editArray(int[] input, int x) {

		int lCount = 0;
		for (int i = 0, j = -1, k = 0; i < input.length; i++) {

			if (input[i] == x) {

				lCount++;
				if (lCount > 2) {
					if (j == -1) {
						j = i;
					}
					input[i] = 0;
				}
			} else if (lCount > 0) {

				if (j != -1) {
					k = j;
					j = -1;
				}
				input[k++] = input[i];
				input[i] = 0;
			}

		}
	}

/**
	 * 
	 * @param input Array of positive numbers in the sorted order.
	 * @param x 
	 */
	private static void editArray(int[] input, int x) {

		int lCount = 0;
		for (int i = 0, j = -1, k = 0; i < input.length; i++) {

			if (input[i] == x) {

				lCount++;
				if (lCount > 2) {
					if (j == -1) {
						j = i;
					}
					input[i] = 0;
				}
			} else if (lCount > 0) {

				if (j != -1) {
					k = j;
					j = -1;
				}
				input[k++] = input[i];
				input[i] = 0;
			}

		}
	}

def editArray(a, m):

    cur_el = None
    cur_el_count = 0
    cur = 0
    
    for i in range(0, len(a)):
        if a[i] != cur_el:
            cur_el_count = 0
            cur_el = a[i]

        if a[i] == cur_el:
            cur_el_count += 1

            if cur_el_count <= min(m, 2):
                a[cur] = a[i]
                cur += 1

    for i in range(cur, len(a)):
        a[i] = 0

Javascript solution

var arr = [1,2,2,2,2,3,4];
var m = 4;
var cur = 0, i, currElement = -1;
var count = 0;

console.log(arr)
for(var i = 0;i < arr.length;i++)
{

  // set the current element
  if (currElement != arr[i])
  {
    currElement = arr[i];
    count = 0;
  }

  // the moment count is > than min, it wil stop incrementing the "curr" pointer
  // and wait for next condition to match
  // so that subsequent filling will resume and start removing the duplicates.
  if (currElement == arr[i] && count < Math.min(m, 2))
  {
    count++;
    // start filling the array with matching value
    arr[cur] = arr[i];
    cur++;
  }
}

for(var i = cur;i < arr.length;i++) {
  arr[i] = 0;
}
console.log(arr)

Java Solution

public static void editArray(int[] arr, int m) {
		int curElm = 0, curIndx = 0, count = 0;
		for(int i = 0; i < arr.length; i++) {
			if(curElm == arr[i]) {
				count++;
			} else count = 1;
			
			if(count <= Math.min(2,  m)) {
				arr[curIndx] = arr[i];
				curElm = arr[i];
				curIndx++;
			}
		}
		
		while(curIndx < arr.length) {
			arr[curIndx] = 0;
			curIndx++;
		}
	}

void editArray(vector<int>& arr, int m)
{
	auto p0 = arr.begin();
	auto p1 = arr.begin();
	while (p1 != arr.end())
	{
		auto er = equal_range(p1, arr.end(), *p1);
		auto p2 = er.second;
		int sz = distance(p1, p2);
		if (sz == m)
			sz = min(m, 2);

		p0 = copy(p1, p1 + sz, p0);
		p1 = p2;
	}

	arr.erase(p0, arr.end());
}

private static void EditArray(int[] arr, int x)
        {
            int lCount = 0;
            for (int i = 0, j = -1, k = 0; i < arr.Length; i++)
            {
                if (arr[i] == x)
                {
                    lCount++;
                    if (lCount > 2)
                    {
                        if (j == -1)//Set the position of i starting where lCount>2, i.e. we are going to assign the 0 to arr[i]
                        {
                            j = i;
                        }
                        arr[i] = 0;
                    }
                }
                else if (lCount > 0)//if x is found more than x
                {
                    if (j != -1)//if j has been assigned the value of i, i.e. position where and on we have set arr[i] to 0  
                    {
                        k = j;
                        j = -1;
                    }
                    if (k > 0)
                    {
                        arr[k++] = arr[i];
                        arr[i] = 0;
                    }
                }
            }

}