Given an array A[] of N integers, the duty is to search out the minimal variety of operations to type the array in non-decreasing order, by selecting an integer X and changing all of the occurrences of X within the array with 0.
Examples:
Enter: N = 5, A[] = {2, 2, 1, 1, 3}
Output: 1
Rationalization: We select X = 2 and substitute all of the occurrences of two with 0. Now the array turns into {2, 2, 1, 1, 3} -> {0, 0, 1, 1, 3} , which is sorted in growing order.Enter: N = 4, A[] = {2, 4, 1, 2}
Output: 3
Method: The issue will be solved simply with the assistance of a Map.
Observations:
There are 2 instances that must be thought of :
- Case 1: Similar factor happens greater than as soon as non-contiguously
- Contemplate the array : {1,6,3,4,5,3,2}.
- Now, since 3 at index 5 is larger than its subsequent factor, so we are going to make that 0 (in addition to 3 at index 2).
- The array turns into {1,6,0,4,5,0,2}.
- So, the one strategy to type the array could be to make all the weather earlier than the zeroes equal to 0. i.e. the array turns into {0,0,0,0,0,0,2}.
- Case 2: Aspect at ith index is larger than the factor at (i+1)th index :
- Contemplate the array : {1,2,3,5,4}.
- Because the factor on the third index is larger than the factor at 4th index, we now have to make the factor at third index equal to zero.
- So , the array turns into {1,2,3,0,4}.
- Now, the one strategy to type the array could be to make all the weather earlier than the zero equal to 0. i.e. the array turns into {0,0,0,0,4}.
It may be noticed that ultimately Case 2 breaks all the way down to Case 1.
Contemplating the above instances, the issue will be solved following the beneath steps :
- Declare a hash map and add the frequency of every factor of the array into the map.
- Iterate by the array from the again, i.e. from i=N-1 to i=0.
- At every iteration, deal with Instances 1 and a couple of as defined above.
- If iteration completes, return 0.
Beneath is the implementation of this method:
C++
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Time Complexity: O(N)
Auxiliary House: O(N)
