Minimum increments or decrements by 1 required to make all array elements in AP - GeeksforGeeks

07 Sep, 2021
Given an array arr[] consisting of N integers, the task is to find the minimum number of increment/decrement by 1 required to be performed on array elements to make all the elements of the given array arr[] in AP. If it is not possible to make the array in AP, then print “-1”.
Examples:
Input: arr[] = {19, 16, 9, 5, 0}Output: 3Explanation:Following are the order of increment/decrements of array elements is made:
Increment array element arr[0](= 19) by 1.Decrement array element arr[1](= 16) by 1.Increment array element arr[2](= 9) by 1.
Increment array element arr[0](= 19) by 1.
Decrement array element arr[1](= 16) by 1.
Increment array element arr[2](= 9) by 1.
After the above operations, the array arr[] modifies to {20, 15, 10, 5, 0}, which is in AP with first term 20 and common difference -5. Therefore, the total count of element is 3.
Input: arr[] = {1, 2, 3, 4, 10}Output: -1
Approach: The given problem can be solved by finding the first term and the common difference from the first two elements and then check if all elements can be changed to the AP sequence with the given first term and common difference by simply iterating over the array. Follow the steps below to solve the problem: 
If N is less than equal to 2, then print 0, because each such sequence is an Arithmetic Progression.
Initialize a variable say, res as N + 1 to store the answer.
Iterate in the range [-1, 1] using the variable a and perform the following steps: Iterate in the range [-1, 1] using the variable b and perform the following steps:Initialize a variable say changes as 0 to store the count of changed elements of the array arr[].If a is not equal to 0, then increase the value of changes by 1.If b is not equal to 0, then increase the value of changes by 1.Initialize a variable say, orig as arr[0] + a to store the first element and diff as (arr[1] + b) – (arr[0] + a) to store the common difference of the arithmetic progression.Initialize a variable say, good as true to store whether the arithmetic progression sequence with first term orig and common difference diff is possible or not.Iterate in the range  [2, N-1] using the variable i and perform the following steps:Initialize a variable actual as orig+i*diff to store the actual element of the arithmetic progression at index i.If abs(actual – arr[i]) is greater than 1, then such arithmetic progression is unreachable. Then Set the value of good to false and break the loop.Else, if actual is not equal to arr[i], increase the value of changes by 1.After iterating through the inner for loop, update the value of res to min(changes, res).
Iterate in the range [-1, 1] using the variable b and perform the following steps:Initialize a variable say changes as 0 to store the count of changed elements of the array arr[].If a is not equal to 0, then increase the value of changes by 1.If b is not equal to 0, then increase the value of changes by 1.Initialize a variable say, orig as arr[0] + a to store the first element and diff as (arr[1] + b) – (arr[0] + a) to store the common difference of the arithmetic progression.Initialize a variable say, good as true to store whether the arithmetic progression sequence with first term orig and common difference diff is possible or not.Iterate in the range  [2, N-1] using the variable i and perform the following steps:Initialize a variable actual as orig+i*diff to store the actual element of the arithmetic progression at index i.If abs(actual – arr[i]) is greater than 1, then such arithmetic progression is unreachable. Then Set the value of good to false and break the loop.Else, if actual is not equal to arr[i], increase the value of changes by 1.After iterating through the inner for loop, update the value of res to min(changes, res).
Initialize a variable say changes as 0 to store the count of changed elements of the array arr[].
If a is not equal to 0, then increase the value of changes by 1.
If b is not equal to 0, then increase the value of changes by 1.
Initialize a variable say, orig as arr[0] + a to store the first element and diff as (arr[1] + b) – (arr[0] + a) to store the common difference of the arithmetic progression.
Initialize a variable say, good as true to store whether the arithmetic progression sequence with first term orig and common difference diff is possible or not.
Iterate in the range  [2, N-1] using the variable i and perform the following steps:Initialize a variable actual as orig+i*diff to store the actual element of the arithmetic progression at index i.If abs(actual – arr[i]) is greater than 1, then such arithmetic progression is unreachable. Then Set the value of good to false and break the loop.Else, if actual is not equal to arr[i], increase the value of changes by 1.
Initialize a variable actual as orig+i*diff to store the actual element of the arithmetic progression at index i.
If abs(actual – arr[i]) is greater than 1, then such arithmetic progression is unreachable. Then Set the value of good to false and break the loop.
Else, if actual is not equal to arr[i], increase the value of changes by 1.
After iterating through the inner for loop, update the value of res to min(changes, res).
After completing the above steps, if res is greater than N then assign -1 to res. Otherwise, print the value of res as the answer.
Below is the implementation of the above approach:
C++
Java
Python3
C#
Javascript
// C++ program for the above approach #include <bits/stdc++.h>using namespace std; // Function to find the minimum number// of elements to be changed to convert// the array into an APint findMinimumMoves(int N, int arr[]){    // If N is less than 3    if (N <= 2) {        return 0;    }     // Stores the answer    int res = N + 1;     // Iterate in the range [-1, 1]    for (int a = -1; a <= 1; a++) {         // Iterate in the range [-1, 1]        for (int b = -1; b <= 1; b++) {             // Stores the total changes            int changes = 0;             if (a != 0) {                changes++;            }             if (b != 0) {                changes++;            }             // Stores the first element            // of the AP            int orig = (arr[0] + a);             // Stores the common difference            // of the AP            int diff = (arr[1] + b) - (arr[0] + a);             // Stores whether it is            // possible to convert the            // array into AP            bool good = true;             // Iterate in the range [2, N-1]            for (int i = 2; i < N; i++) {                 // Stores the ith element                // of the AP                int actual = orig + i * diff;                 // If abs(actual-arr[i])                // is greater than 1                if (abs(actual - arr[i]) > 1) {                     // Mark as false                    good = false;                    break;                }                // If actual is not                // equal to arr[i]                if (actual != arr[i])                    changes++;            }            if (!good)                continue;             // Update the value of res            res = min(res, changes);        }    }     // If res is greater than N    if (res > N)        res = -1;     // Return the value of res    return res;} // Driver Codeint main(){    int arr[] = { 19, 16, 9, 5, 0 };    int N = sizeof(arr) / sizeof(arr[0]);    cout << findMinimumMoves(N, arr);     return 0;}
// Java program for the above approachimport java.io.*; class GFG{     // Function to find the minimum number// of elements to be changed to convert// the array into an APstatic int findMinimumMoves(int N, int arr[]){         // If N is less than 3    if (N <= 2)    {        return 0;    }     // Stores the answer    int res = N + 1;     // Iterate in the range [-1, 1]    for(int a = -1; a <= 1; a++)    {                 // Iterate in the range [-1, 1]        for(int b = -1; b <= 1; b++)        {                         // Stores the total changes            int changes = 0;             if (a != 0)            {                changes++;            }             if (b != 0)            {                changes++;            }             // Stores the first element            // of the AP            int orig = (arr[0] + a);             // Stores the common difference            // of the AP            int diff = (arr[1] + b) - (arr[0] + a);             // Stores whether it is            // possible to convert the            // array into AP            boolean good = true;             // Iterate in the range [2, N-1]            for(int i = 2; i < N; i++)            {                                 // Stores the ith element                // of the AP                int actual = orig + i * diff;                 // If abs(actual-arr[i])                // is greater than 1                if (Math.abs(actual - arr[i]) > 1)                {                                         // Mark as false                    good = false;                    break;                }                                 // If actual is not                // equal to arr[i]                if (actual != arr[i])                    changes++;            }            if (!good)                continue;             // Update the value of res            res = Math.min(res, changes);        }    }     // If res is greater than N    if (res > N)        res = -1;     // Return the value of res    return res;} // Driver Codepublic static void main(String[] args){    int arr[] = { 19, 16, 9, 5, 0 };    int N = arr.length;     System.out.println(findMinimumMoves(N, arr));}} // This code is contributed by Potta Lokesh
# Python3 program for the above approach # Function to find the minimum number# of elements to be changed to convert# the array into an APdef findMinimumMoves(N, arr):         # If N is less than 3    if (N <= 2):        return 0     # Stores the answer    res = N + 1     # Iterate in the range [-1, 1]    for a in range(-1, 2, 1):         # Iterate in the range [-1, 1]        for b in range(-1, 2, 1):                         # Stores the total changes            changes = 0             if (a != 0):                changes += 1             if (b != 0):                changes += 1             # Stores the first element            # of the AP            orig = (arr[0] + a)             # Stores the common difference            # of the AP            diff = (arr[1] + b) - (arr[0] + a)             # Stores whether it is            # possible to convert the            # array into AP            good = True             # Iterate in the range [2, N-1]            for i in range(2, N, 1):                                 # Stores the ith element                # of the AP                actual = orig + i * diff                 # If abs(actual-arr[i])                # is greater than 1                if (abs(actual - arr[i]) > 1):                     # Mark as false                    good = False                    break                 # If actual is not                # equal to arr[i]                if (actual != arr[i]):                    changes += 1             if (good == False):                continue             # Update the value of res            res = min(res, changes)     # If res is greater than N    if (res > N):        res = -1     # Return the value of res    return res # Driver Codeif __name__ == '__main__':         arr = [ 19, 16, 9, 5, 0 ]    N = len(arr)         print(findMinimumMoves(N, arr)) # This code is contributed by ipg2016107
// C# program for the above approachusing System; class GFG{     // Function to find the minimum number// of elements to be changed to convert// the array into an APstatic int findMinimumMoves(int N, int[] arr){         // If N is less than 3    if (N <= 2)    {        return 0;    }     // Stores the answer    int res = N + 1;     // Iterate in the range [-1, 1]    for(int a = -1; a <= 1; a++)    {                 // Iterate in the range [-1, 1]        for(int b = -1; b <= 1; b++)        {                         // Stores the total changes            int changes = 0;             if (a != 0)            {                changes++;            }             if (b != 0)            {                changes++;            }             // Stores the first element            // of the AP            int orig = (arr[0] + a);             // Stores the common difference            // of the AP            int diff = (arr[1] + b) - (arr[0] + a);             // Stores whether it is            // possible to convert the            // array into AP            bool good = true;             // Iterate in the range [2, N-1]            for(int i = 2; i < N; i++)            {                                 // Stores the ith element                // of the AP                int actual = orig + i * diff;                 // If abs(actual-arr[i])                // is greater than 1                if (Math.Abs(actual - arr[i]) > 1)                {                                         // Mark as false                    good = false;                    break;                }                                 // If actual is not                // equal to arr[i]                if (actual != arr[i])                    changes++;            }            if (!good)                continue;             // Update the value of res            res = Math.Min(res, changes);        }    }     // If res is greater than N    if (res > N)        res = -1;     // Return the value of res    return res;}  // Driver codestatic public void Main(){    int[] arr = { 19, 16, 9, 5, 0 };    int N = arr.Length;     Console.WriteLine(findMinimumMoves(N, arr));}} // This code is contributed by target_2.
<script> // Javascript program for the above approach // Function to find the minimum number// of elements to be changed to convert// the array into an APfunction findMinimumMoves(N, arr){         // If N is less than 3    if (N <= 2)    {        return 0;    }     // Stores the answer    let res = N + 1;     // Iterate in the range [-1, 1]    for(let a = -1; a <= 1; a++)    {                 // Iterate in the range [-1, 1]        for(let b = -1; b <= 1; b++)        {                         // Stores the total changes            let changes = 0;             if (a != 0)            {                changes++;            }             if (b != 0)            {                changes++;            }             // Stores the first element            // of the AP            let orig = (arr[0] + a);             // Stores the common difference            // of the AP            let diff = (arr[1] + b) - (arr[0] + a);             // Stores whether it is            // possible to convert the            // array into AP            let good = true;             // Iterate in the range [2, N-1]            for(let i = 2; i < N; i++)            {                                 // Stores the ith element                // of the AP                let actual = orig + i * diff;                 // If abs(actual-arr[i])                // is greater than 1                if (Math.abs(actual - arr[i]) > 1)                {                                         // Mark as false                    good = false;                    break;                }                                 // If actual is not                // equal to arr[i]                if (actual != arr[i])                    changes++;            }            if (!good)                continue;             // Update the value of res            res = Math.min(res, changes);        }    }     // If res is greater than N    if (res > N)        res = -1;     // Return the value of res    return res;} // Driver Codelet arr = [ 19, 16, 9, 5, 0 ];let N = arr.length; document.write(findMinimumMoves(N, arr)); // This code is contributed by target_2     </script>
3
 
Time Complexity: O(N)Auxiliary Space: O(1)
lokeshpotta20
target_2
ipg2016107
khushboogoyal499
arithmetic progression
Arithmetic Progressions
Arrays
Mathematical
Sorting
Arrays
Mathematical
Sorting
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