Given an array of integer numbers, we need to sort this array in a minimum number of steps where in one step we can insert any array element from its position to any other position.

Examples:

Input : arr[] = [2, 3, 5, 1, 4, 7, 6] Output : 3 We can sort above array in 3 insertion steps as shown below, 1 before array value 2 4 before array value 5 6 before array value 7 Input : arr[] = {4, 6, 5, 1} Output : 2

We can solve this problem using dynamic programming. The main thing to observe is that moving an element doesn’t change the relative order of elements other than the element which is being moved. Now consider longest increasing subsequenc (LIS)e in which equal element are also taken as part of the increasing sequence, now if keep the element of this increasing sequence as it is and move all other elements then it will take the least number of steps because we have taken longest subsequence which does not need to be moved. Finally, the answer will be the size of the array minus the size of the longest increasing subsequence.

As LIS problem can be solved in O(N^2) with O(N) extra space using Dynamic Programming.

Below is C++ implementation of above idea.

## C++

// C++ program to get minimum number of insertion // steps to sort an array #include <bits/stdc++.h> using namespace std; // method returns min steps of insertion we need // to perform to sort array 'arr' int minInsertionStepToSortArray(int arr[], int N) { // lis[i] is going to store length of lis // that ends with i. int lis[N]; /* Initialize lis values for all indexes */ for (int i = 0; i < N; i++) lis[i] = 1; /* Compute optimized lis values in bottom up manner */ for (int i = 1; i < N; i++) for (int j = 0; j < i; j++) if (arr[i] >= arr[j] && lis[i] < lis[j] + 1) lis[i] = lis[j] + 1; /* The overall LIS must end with of of the array elements. Pick maximum of all lis values */ int max = 0; for (int i = 0; i < N; i++) if (max < lis[i]) max = lis[i]; // return size of array minus length of LIS // as final result return (N - max); } // Driver code to test above methods int main() { int arr[] = {2, 3, 5, 1, 4, 7, 6}; int N = sizeof(arr) / sizeof(arr[0]); cout << minInsertionStepToSortArray(arr, N); return 0; }

## Java

// Java program to get minimum number of insertion // steps to sort an array class Main { // method returns min steps of insertion we need // to perform to sort array 'arr' static int minInsertionStepToSortArray(int arr[], int N) { // lis[i] is going to store length of lis // that ends with i. int[] lis = new int[N]; /* Initialize lis values for all indexes */ for (int i = 0; i < N; i++) lis[i] = 1; /* Compute optimized lis values in bottom up manner */ for (int i = 1; i < N; i++) for (int j = 0; j < i; j++) if (arr[i] >= arr[j] && lis[i] < lis[j] + 1) lis[i] = lis[j] + 1; /* The overall LIS must end with of of the array elements. Pick maximum of all lis values */ int max = 0; for (int i = 0; i < N; i++) if (max < lis[i]) max = lis[i]; // return size of array minus length of LIS // as final result return (N - max); } // Driver code to test above methods public static void main (String[] args) { int arr[] = {2, 3, 5, 1, 4, 7, 6}; int N = arr.length; System.out.println(minInsertionStepToSortArray(arr, N)); } } /* This code is contributed by Harsh Agarwal */

Output:

3

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