# Minimum number of steps to make all elements of array equal

Given two arrays A[] and B[] of the same length, the task is to find the minimum number of steps required to make all the elements of the array equal by replacing the element to the difference of the corresponding element from array B.

Note: If this is not possible print -1.

Examples:

Input: A[] = {5, 7, 10, 5, 15} B[] = {2, 2, 1, 3, 5}
Output: 8
Explanation:
Steps to convert Array elements equal –
Index 0: 5 is not changed, Steps Taken = 0
Index 1: 7 – (2 * 1) = 5, Steps Taken = 1
Index 2: 10 – (1 * 5) = 5, Steps Taken = 5
Index 3: 5 is not changed, Steps Taken = 0
Index 4: 15 – (5 * 2) = 5, Steps Taken = 2
Therefore, total steps required = 0 + 1 + 5 + 0 + 2 = 8

Input: A[] = {5, 6}, B[] = {4, 3}
Output: -1
Explanation:
It’s not possible to make all elements of array equal.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea is to find the value at which the array elements can be converged, this value will definitely lie in between 0 to the minimum value of the array. Below is the illustration of approach:

• Intially, mark flag as False, to check that the array is convertible or not.
• Run a while loop until the minimum of the array is greater than -1.
• Iterate over the indices of the array, that is from 0 to length – 1
• For each index, check that if the element at that index in array A is greater than the minimum of the array, If yes then update the array element A[i] as A[i] – B[i].
• Increment the number of steps by 1.
• Check that all the array elements are equal if yes then mark the flag as True and break the while loop
• Otherwise, repeat the steps above to compute the total number of steps required.
• Finally, check if the flag is True, If yes then the array is convertible.

Explanation with Example:
Given arrays be – A[] = {5, 7, 10, 5, 15}, B[] = {2, 2, 1, 3, 5}

Array A Current Index Minimum of Array A Steps Comments
{5, 7, 10, 5, 15} 0 5 0 No operation !
{5, 5, 10, 5, 15} 1 5 1 Array element is subtracted once. 7 – 2 = 5
{5, 5, 9, 5, 15} 2 5 2 Array element is subtracted once. 10 – 1 = 9
{5, 5, 8, 5, 15} 2 5 3 Array element is subtracted once. 9 – 1 = 8
{5, 5, 7, 5, 15} 2 5 4 Array element is subtracted once. 8 – 1 = 7
{5, 5, 6, 5, 15} 2 5 5 Array element is subtracted once. 7 – 1 = 6
{5, 5, 5, 5, 15} 2 5 6 Array element is subtracted once. 6 – 1 = 5
{5, 5, 5, 5, 10} 4 5 7 Array element is subtracted once. 15 – 5 = 10
{5, 5, 5, 5, 5} 4 5 8 Array element is subtracted once. 10 – 5 = 5

Below is the implementation of the above approach:

## Python

 `# Python implementation to find the ` `# minimum number of steps to convert ` `# array by subtracting the corresponding ` `# element from array B ` ` `  `# Function to find the minimum steps ` `def` `minimumSteps(ar1, ar2, n): ` `     `  `    ``# Counter to store the steps ` `    ``ans ``=` `0` `     `  `    ``# Flag to check that  ` `    ``# array is converted ` `    ``flag ``=` `True` `     `  `    ``# Loop until the minimum of the  ` `    ``# array is greater than -1 ` `    ``while` `min``(ar1)>``-``1``: ` `        ``a ``=` `min``(ar1) ` `         `  `        ``# Loop to convert the array  ` `        ``# elements by substraction ` `        ``for` `i ``in` `range``(n): ` `            ``if` `ar1[i]!``=` `a: ` `                ``ar1[i]``-``=` `ar2[i] ` `                ``ans``+``=` `1` `                 `  `        ``# If the array is converted ` `        ``if` `len``(``set``(ar1))``=``=` `1``: ` `            ``flag ``=` `False` `            ``print``(ans) ` `            ``break` `    ``if` `flag: ` `        ``print``(``-``1``) ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` `    ``n ``=` `5` `    ``ar1 ``=` `[``5``, ``7``, ``10``, ``5``, ``15``] ` `    ``ar2 ``=` `[``1``, ``2``, ``1``, ``5``, ``5``] ` `     `  `    ``# Function Call ` `    ``minimumSteps(ar1, ar2, n) `

Output:

```8
```

Performance Analysis:

• Time Complexity: As in the above approach, there are two loops to convert the array which takes O(N2) in worst case, Hence the Time Complexity will be O(N2).
• Space Complexity: As in the above approach, there is no extra space used, Hence the space complexity will be O(1).

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