Minimum steps for increasing and decreasing Array to reach either 0 or N

Given an integer N and two arrays increasing[] and decreasing[], such that they have elements from 1 to N only. The task is to find the minimum number of steps for each element of the two arrays to reach either 0 or N. A step is defined as follows:

• In one step, all the elements of the increasing[] array increases by 1, and all the elements of the decreasing[] array decreases by 1.
• When an element becomes either 0 or N, no more increase or decrease operation is performed on it.

Examples:

Input: N = 5, increasing[] = {1, 2}, decreasing[] = {3, 4} 
Output: 4
Explanation:
Step 1: increasing[] array becomes {2, 3}, decreasing[] = {2, 3} 
Step 2: increasing[] array becomes {3, 4}, decreasing[] = {1, 2} 
Step 3: increasing[] array becomes {4, 5}, decreasing[] = {0, 1} 
Step 4: increasing[] array becomes {5, 5}, decreasing[] = {0, 0} 
4 Steps are required for all elements to become either 0 or N. Hence, the output is 4.

Input: N = 7, increasing[] = {3, 5}, decreasing[] = {6} 
Output: 6

Approach: The idea is to find the maximum between the steps required by all the elements of the increasing[] array and the decreasing[] array to reach N and 0 respectively. Below are the steps:

1. Find the minimum element of the array increasing[].
2. The maximum steps taken by all the elements of the increasing[] array to reach N is given by N – min(increasing[]).
3. Find the maximum element of the array decreasing[].
4. The maximum steps taken by all the elements of the decreasing[] array to reach 0 is given by max(decreasing[]).
5. Therefore, the minimum number of steps when all the elements become either 0 or N is given by max(N – min(increasing[]), max(decreasing[])).

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function that finds the minimum
// steps to reach either 0 or N for
// given increasing and decreasing
// arrays
void minSteps(int N, int increasing[],
              int decreasing[], int m1, int m2)
{

    // Initialize variable to
    // find the minimum element
    int mini = INT_MAX;

    // Find minimum element in
    // increasing[] array
    for(int i = 0; i < m1; i++)
    {
        if (mini > increasing[i])
            mini = increasing[i];
    }

    // Initialize variable to
    // find the maximum element
    int maxi = INT_MIN;

    // Find maximum element in
    // decreasing[] array
    for(int i = 0; i < m2; i++)
    {
        if (maxi < decreasing[i])
            maxi = decreasing[i];
    }

    // Find the minimum steps
    int minSteps = max(maxi,
                       N - mini);

    // Print the minimum steps
    cout << minSteps << endl;
}

// Driver code
int main()
{

    // Given N
    int N = 7;

    // Given increasing
    // and decreasing array
    int increasing[] = { 3, 5 };
    int decreasing[] = { 6 };

    // Find length of arrays
    // increasing and decreasing
    int m1 = sizeof(increasing) /sizeof(increasing[0]);
    int m2 = sizeof(decreasing) / sizeof(decreasing[0]);

    // Function call
    minSteps(N, increasing, decreasing, m1, m2);
}

// This code is contributed by Manne Sree Charan

Java

// Java program for the above approach
import java.util.*;

public class GFG {

    // Function that finds the minimum
    // steps to reach either 0 or N for
    // given increasing and decreasing
    // arrays
    public static void
    minSteps(int N, int[] increasing,
            int[] decreasing)
    {

        // Initialize variable to
        // find the minimum element
        int min = Integer.MAX_VALUE;

        // Find minimum element in
        // increasing[] array
        for (int i : increasing) {
            if (min > i)
                min = i;
        }

        // Initialize variable to
        // find the maximum element
        int max = Integer.MIN_VALUE;

        // Find maximum element in
        // decreasing[] array
        for (int i : decreasing) {
            if (max < i)
                max = i;
        }

        // Find the minimum steps
        int minSteps = Math.max(max,
                                N - min);

        // Print the minimum steps
        System.out.println(minSteps);
    }

    // Driver Code
    public static void main(String[] args)
    {
        // Given N
        int N = 7;

        // Given increasing
        // and decreasing array
        int increasing[] = { 3, 5 };
        int decreasing[] = { 6 };

        // Function call
        minSteps(N, increasing, decreasing);
    }
}

Python3

# Python3 program for
# the above approach
import sys

# Function that finds the minimum
# steps to reach either 0 or N for
# given increasing and decreasing
# arrays
def minSteps(N, increasing, decreasing):
    # Initialize variable to
    # find the minimum element
    Min = sys.maxsize;

    # Find minimum element in
    # increasing array
    for i in increasing:
        if (Min > i):
            Min = i;

    # Initialize variable to
    # find the maximum element
    Max = -sys.maxsize;

    # Find maximum element in
    # decreasing array
    for i in decreasing:
        if (Max < i):
            Max = i;

    # Find the minimum steps
    minSteps = max(Max, N - Min);

    # Print the minimum steps
    print(minSteps);

# Driver Code
if __name__ == '__main__':

    # Given N
    N = 7;

    # Given increasing
    # and decreasing array
    increasing = [3, 5];
    decreasing = [6];

    # Function call
    minSteps(N, increasing, decreasing);

# This code contributed by Rajput-Ji

C#

// C# program for the above approach
using System;

class GFG{

// Function that finds the minimum
// steps to reach either 0 or N for
// given increasing and decreasing
// arrays
public static void minSteps(int N, int[] increasing,
                                   int[] decreasing)
{

    // Initialize variable to
    // find the minimum element
    int min = int.MaxValue;

    // Find minimum element in
    // increasing[] array
    foreach(int i in increasing)
    {
        if (min > i)
            min = i;
    }

    // Initialize variable to
    // find the maximum element
    int max = int.MinValue;

    // Find maximum element in
    // decreasing[] array
    foreach(int i in decreasing)
    {
        if (max < i)
            max = i;
    }

    // Find the minimum steps
    int minSteps = Math.Max(max,
                            N - min);

    // Print the minimum steps
    Console.WriteLine(minSteps);
}

// Driver Code
public static void Main(String[] args)
{

    // Given N
    int N = 7;

    // Given increasing
    // and decreasing array
    int []increasing = { 3, 5 };
    int []decreasing = { 6 };

    // Function call
    minSteps(N, increasing, decreasing);
}
}

// This code is contributed by Amit Katiyar

Javascript

<script>

// Javascript program for the above approach

// Function that finds the minimum
// steps to reach either 0 or N for
// given increasing and decreasing
// arrays
function minSteps(N, increasing, decreasing, m1, m2)
{

    // Initialize variable to
    // find the minimum element
    var mini = 2147483647;

    var i;
    // Find minimum element in
    // increasing[] array
    for(i = 0; i < m1; i++)
    {
        if (mini > increasing[i])
            mini = increasing[i];
    }

    // Initialize variable to
    // find the maximum element
    var maxi = -2147483648;

    // Find maximum element in
    // decreasing[] array
    for(i = 0; i < m2; i++)
    {
        if (maxi < decreasing[i])
            maxi = decreasing[i];
    }

    // Find the minimum steps
    var minSteps = Math.max(maxi,N - mini);

    // Print the minimum steps
    document.write(minSteps);
}

// Driver code

    // Given N
    var N = 7;

    // Given increasing
    // and decreasing array
    var increasing = [3, 5];
    var decreasing = [6];

    // Find length of arrays
    // increasing and decreasing
    var m1 = increasing.length;
    var m2 = decreasing.length;

    // Function call
    minSteps(N, increasing, decreasing, m1, m2);

// This code is contributed by bgangwar59.
</script>

Output:

6

Time Complexity: O(N)
Auxiliary Space: O(1)