CoCubes Programming Question – 2

November 17, 2020

Maximum difference between two elements such that larger element appears after the smaller number

Given an array arr[] of integers, find out the difference between any two elements such that larger element appears after the smaller number in arr[].

Examples: If array is [2, 3, 10, 6, 4, 8, 1] then returned value should be 8 (Diff between 10 and 2). If array is [ 7, 9, 5, 6, 3, 2 ] then returned value should be 2 (Diff between 7 and 9)

Time Complexity: O(n^2)
Auxiliary Space: O(1)

Use two loops. In the outer loop, pick elements one by one and in the inner loop calculate the difference of the picked element with every other element in the array and compare the difference with the maximum difference calculated so far.

31 comments on “CoCubes Programming Question – 2”


  • sonali

    import java.util.*;
    public class MyClass {
    public static void main(String args[]) {
    int []arr={7,9,5,6,3,2};
    int n=arr.length-1;
    ArrayList arr1=new ArrayList();
    for(int i=0;i<n-1;i++)
    {
    for(int j=i+1;j<n;j++)
    {
    if(arr[i]<arr[j])
    arr1.add(arr[j]-arr[i]);
    }
    }
    int max=Collections.max(arr1);
    System.out.println(max);
    }
    }

    0

  • Pavan

    arr=list(map(int,input().split()))
    n=len(arr)
    diff=0
    for i in range(n):
    for j in range(i+1,n):
    if arr[i]<arr[j]:
    if diff<(arr[j]-arr[i]):
    diff=arr[j]-arr[i]
    print(diff)

    0

  • Amarnath

    import java.util.*;
    public class Main
    {
    public static void main(String[] args) {
    int[] arr=new int[]{7, 9, 5, 6, 3, 2};
    Arrays.sort(arr);
    System.out.println(arr[arr.length-1]-arr[arr.length-2]);
    }
    }

    0

    • Anjana

      this wouldn’t work as by sorting we are disturbing the order and in the question we have constraint as “such that larger element appears after the smaller number “.

      0

  • Ravi Chandra

    n=int(input(“Enter the size of the array”))
    array=list(map(int,input().split()))
    ma=array[0]
    for i in range(n):
    for j in range(i+1,n):
    if(array[i]ma):
    ma=array[j]-array[i]
    print(ma)

    1

  • SaiManoj

    #include
    int main()
    {
    int n,a[100],i,dif,max=0;
    scanf(“%d”,&n);
    for(i=1;i<=n;i++)
    scanf("%d",&a[i]);
    dif=a[1]-a[0];
    for(i=1;i<=n;i++)
    {
    for(int j=i+1;j=a[i])
    dif=a[j]-a[i];
    if(max<dif)
    max=dif;
    }
    }
    printf("%d",max);
    return 0;
    }

    0

  • shivam mishra

    l=list(map(int,input().strip().split(” “)))
    l1=[]
    for i in range(len(l)):
    for j in range(i,len(l)):
    if i!=j:
    if l[i]<l[j]:
    l1.append(l[j]-l[i])
    print(max(l1))

    0

  • Mohd Saif

    #include
    using namespace std;
    void diff_arr(int arr[],int n){
    vector temp_arr;
    temp_arr.push_back(arr[0]);
    int k=0;
    for(int i=1;itemp_arr.back()){
    temp_arr.push_back(arr[i]);
    k+=1;
    }
    }
    int x=temp_arr[k];
    int pos2;
    for(int i=0;i<n;i++){
    if(x==arr[i]){
    pos2=i;
    }
    }
    for(int i=1;iarr[i]){
    int pos3=i;
    if(pos3<pos2){
    temp_arr[0]=arr[i];
    }
    }
    }
    x=(temp_arr[k]-temp_arr[0]);
    cout<>n;
    int arr[n];
    for(int i=0;i>arr[i];
    }
    diff_arr(arr,n);
    return 0;
    }

    0

  • prince

    arr=list(map(int,input().split()))
    arr1=arr.copy()
    arr1.sort()
    a=arr1[len(arr1)-1]
    b=arr.index(a)
    mylist=list(arr[0:b])
    mylist.sort()
    c=mylist[0]
    print (a-c)

    0

  • Suraj Potti

    Solution for this question in Python.
    maxi = 0
    diff = 0
    arr = [ 9, 7, 6, 5, 3, 2 ]
    for i in range(len(arr)-1):
    sub_max = max(arr[i:len(arr)]) #slicing the array after the current index.
    if(i maxi):
    maxi = diff
    else:
    pass
    else:
    pass
    print(maxi)
    This code will return zero if no such numbers satisfying the conditions exists.

    0
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