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Check if empty spaces of Array can be filled maintaining given relations of adjacent

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  • Last Updated : 29 Aug, 2022
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Given two arrays, array A[] of size N and array B[] of size N-1, array A has all the positive elements and -1 for the empty values and array B has 3 characters {‘>’, ‘=’, ‘<‘} which indicates the following:

  1. >‘ means A[i] > A[i+1]
  2. =‘ means A[i] = A[i+1]
  3. <‘ means A[i] < A[i+1]

The task is to find if we can fill -1(empty values) of A with some positive values that also satisfy the relations between the adjacents mentioned in array B.

Examples: 

Input: N = 6, A = {8, 6, -1, -1, -1, 11}, B = {‘>’, ‘>’, ‘=’, ‘<‘, ‘>’}
Output: Yes
Explanation: Consider, A becomes {8, 6, 5, 5, 12, 11},  
It satisfies B, so the answer will be ‘Yes’.

Input: N = 5, A = {4, 1, -1, -1, 7}, B = {‘<‘, ‘<‘, ‘<‘, ‘<‘}
Output: No

Approach: The problem can be solved based on the following mathematical idea:

Use range to compare whether we can have a valid solution for a particular index and change the range accordingly.

Initially, the left range is 0 and the right range is INT_MAX. 

  • If ‘<‘ operator is encountered, then the valid range for A[i+1] is left = A[i]+1 and right = INT_MAX.
  • If ‘>’ operator is encountered, then the valid range for A[i+1] is left = 0 and right = A[i] – 1.
  • If ‘=’ operator is encountered, then the valid range for A[i+1] is left = A[i] and right = A[i]
  • Values of ranges are changed when A[i] is given, but when A[i] = -1 ranges are compared and changed according to the above-mentioned conditions
  • If left ≤ right and left ≥ 0 and right ≥ 0 is satisfied for each index position, then ‘Yes’ is returned, otherwise ‘No’ is returned.

Below is the implementation of the above approach.

C++




// C++ code for this approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check whether array A
// can be made such that all operator
// of array B holds true for each index i
string checkComparisons(int N, vector<int> A,
                        vector<char> B)
{
    // Declaring initial ranges as 0 to Infinity
    int left = 0, right = INT_MAX, index = 0;
    bool valid = 1;
    for (int i = 0; i < N - 1; i++, index++) {
 
        // If current element is not empty
        if (A[i] != -1) {
 
            // Compare normal elements
            // (non empty i.e. != -1) based on B
            if (A[i + 1] != -1) {
                if (B[index] == '>') {
                    if (A[i] <= A[i + 1])
                        return "No";
                }
                if (B[index] == '=') {
                    if (A[i] != A[i + 1])
                        return "No";
                }
                if (B[index] == '<') {
                    if (A[i] >= A[i + 1])
                        return "No";
                }
            }
 
            // Compare and update ranges
            // for empty spaces
            else {
                if (B[index] == '>') {
                    left = 0;
                    right = A[i] - 1;
                }
                if (B[index] == '=') {
                    left = A[i];
                    right = A[i];
                }
                if (B[index] == '<') {
                    left = A[i] + 1;
                    right = INT_MAX;
                }
            }
        }
 
        // If current element is empty
        else {
 
            // Compare and update ranges if
            // next element is not empty
            if (A[i + 1] != -1) {
                if (B[index] == '>') {
                    if (right <= A[i + 1])
                        return "No";
                }
                if (B[index] == '=') {
                    if ((left > A[i + 1])
                        || (right < A[i + 1]))
                        return "No";
                }
                if (B[index] == '<') {
                    if (left >= A[i + 1])
                        return "No";
                }
            }
 
            // Compare and update ranges if
            // next element is empty based on
            // previous ranges
            else {
                if (B[index] == '>') {
                    left = 0;
                    right--;
                }
                if (B[index] == '<') {
                    left++;
                    right = INT_MAX;
                }
            }
        }
        if (left > right)
            return "No";
        if (right < 0)
            return "No";
        if (left < 0)
            return "No";
    }
    return "Yes";
}
 
// Driver Code
int main()
{
    int N = 6;
    vector<int> A{ 8, 6, -1, -1, -1, 11 };
    vector<char> B{ '>', '>', '=', '<', '>' };
 
    // Function call
    cout << checkComparisons(N, A, B);
    return 0;
}


Java




// Java code for this approach
import java.io.*;
 
class GFG {
    // Function to check whether array A
    // can be made such that all operator
    // of array B holds true for each index i
    public static String checkComparisons(int N, int A[],
                                          char B[])
    {
        // Declaring initial ranges as 0 to Infinity
        int left = 0, right = Integer.MAX_VALUE, index = 0;
        boolean valid = true;
        for (int i = 0; i < N - 1; i++, index++) {
 
            // If current element is not empty
            if (A[i] != -1) {
 
                // Compare normal elements
                // (non empty i.e. != -1) based on B
                if (A[i + 1] != -1) {
                    if (B[index] == '>') {
                        if (A[i] <= A[i + 1])
                            return "No";
                    }
                    if (B[index] == '=') {
                        if (A[i] != A[i + 1])
                            return "No";
                    }
                    if (B[index] == '<') {
                        if (A[i] >= A[i + 1])
                            return "No";
                    }
                }
 
                // Compare and update ranges
                // for empty spaces
                else {
                    if (B[index] == '>') {
                        left = 0;
                        right = A[i] - 1;
                    }
                    if (B[index] == '=') {
                        left = A[i];
                        right = A[i];
                    }
                    if (B[index] == '<') {
                        left = A[i] + 1;
                        right = Integer.MAX_VALUE;
                    }
                }
            }
 
            // If current element is empty
            else {
 
                // Compare and update ranges if
                // next element is not empty
                if (A[i + 1] != -1) {
                    if (B[index] == '>') {
                        if (right <= A[i + 1])
                            return "No";
                    }
                    if (B[index] == '=') {
                        if ((left > A[i + 1])
                            || (right < A[i + 1]))
                            return "No";
                    }
                    if (B[index] == '<') {
                        if (left >= A[i + 1])
                            return "No";
                    }
                }
 
                // Compare and update ranges if
                // next element is empty based on
                // previous ranges
                else {
                    if (B[index] == '>') {
                        left = 0;
                        right--;
                    }
                    if (B[index] == '<') {
                        left++;
                        right = Integer.MAX_VALUE;
                    }
                }
            }
            if (left > right)
                return "No";
            if (right < 0)
                return "No";
            if (left < 0)
                return "No";
        }
        return "Yes";
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int N = 6;
        int A[] = { 8, 6, -1, -1, -1, 11 };
        char B[] = { '>', '>', '=', '<', '>' };
 
        // Function call
        System.out.print(checkComparisons(N, A, B));
    }
}
 
// This code is contributed by Rohit Pradhan


Python3




# Python code for this approach
# Function to check whether array A
# can be made such that all operator
# of array B holds true for each index i
import sys
def checkComparisons(N, A, B):
   
  # Declaring initial ranges as 0 to Infinity
  left = 0
  right = -sys.maxsize-1
  index = 0;
  valid = 1;
  for i in range (N-1):
    i = i + 1
    index = index + 1
     
    # If current element is not empty
    if (A[i] != -1):
       
      # Compare normal elements
      # (non empty i.e. != -1) based on B
      if (A[i + 1] != -1):
        if (B[index] == '>'):
          if (A[i] <= A[i + 1]):
            return "No"
        if (B[index] == '='):
          if (A[i] != A[i + 1]):
            return "No"
        if (B[index] == '<'):
          if (A[i] >= A[i + 1]):
            return "No"
           
      # Compare and update ranges
      # for empty spaces
      else:
        if (B[index] == '>'):
          left = 0
          right = A[i] - 1
        if (B[index] == '='):
          left = A[i]
          right = A[i]
        if (B[index] == '<'):
          left = A[i] + 1
          right = -sys.maxsize - 1
           
    # If current element is empty
    else:
       
      # Compare and update ranges if
      # next element is not empty
      if (A[i + 1] != -1):
        if (B[index] == '>'):
          if (right <= A[i + 1]):
            return "No"
        if (B[index] == '='):
          if ((left > A[i + 1]) or (right < A[i + 1])):
            return "No"
        if (B[index] == '<'):
          if (left >= A[i + 1]):
            return "No"
      # Compare and update ranges if
      # next element is empty based on
      # previous ranges
      else:
        if (B[index] == '>'):
          left = 0
          right=right-1
        if (B[index] == '<'):
          left=left+1
          right = -sys.maxsize-1
    if (left > right):
      return "No"
    if (right < 0):
      return "No"
    if (left < 0):
      return "No"
    return "Yes"
   
# Driver Code
N = 6
A = [8, 6, -1, -1, -1, 11]
B = [ '>', '>', '=', '<', '>' ]
 
# Function call
print(checkComparisons(N, A, B))
 
# This code is contributed by Atul_kumar_Shrivastava


C#




// C# implementation of above approach
using System;
using System.Collections.Generic;
 
class GFG {
 
  // Function to check whether array A
  // can be made such that all operator
  // of array B holds true for each index i
  public static String checkComparisons(int N, int[] A,
                                        char[] B)
  {
    // Declaring initial ranges as 0 to Infinity
    int left = 0, right = Int32.MaxValue, index = 0;
    bool valid = true;
    for (int i = 0; i < N - 1; i++, index++) {
 
      // If current element is not empty
      if (A[i] != -1) {
 
        // Compare normal elements
        // (non empty i.e. != -1) based on B
        if (A[i + 1] != -1) {
          if (B[index] == '>') {
            if (A[i] <= A[i + 1])
              return "No";
          }
          if (B[index] == '=') {
            if (A[i] != A[i + 1])
              return "No";
          }
          if (B[index] == '<') {
            if (A[i] >= A[i + 1])
              return "No";
          }
        }
 
        // Compare and update ranges
        // for empty spaces
        else {
          if (B[index] == '>') {
            left = 0;
            right = A[i] - 1;
          }
          if (B[index] == '=') {
            left = A[i];
            right = A[i];
          }
          if (B[index] == '<') {
            left = A[i] + 1;
            right = Int32.MaxValue;
          }
        }
      }
 
      // If current element is empty
      else {
 
        // Compare and update ranges if
        // next element is not empty
        if (A[i + 1] != -1) {
          if (B[index] == '>') {
            if (right <= A[i + 1])
              return "No";
          }
          if (B[index] == '=') {
            if ((left > A[i + 1])
                || (right < A[i + 1]))
              return "No";
          }
          if (B[index] == '<') {
            if (left >= A[i + 1])
              return "No";
          }
        }
 
        // Compare and update ranges if
        // next element is empty based on
        // previous ranges
        else {
          if (B[index] == '>') {
            left = 0;
            right--;
          }
          if (B[index] == '<') {
            left++;
            right = Int32.MaxValue;
          }
        }
      }
      if (left > right)
        return "No";
      if (right < 0)
        return "No";
      if (left < 0)
        return "No";
    }
    return "Yes";
  }
 
  // Driver Code
  public static void Main()
  {
    int N = 6;
    int[] A = { 8, 6, -1, -1, -1, 11 };
    char[] B = { '>', '>', '=', '<', '>' };
 
    // Function call
    Console.WriteLine(checkComparisons(N, A, B));
  }
}
 
// This code is contributed by sanjoy_62.


Javascript




<script>
    // JavaScript code for this approach
 
    const INT_MAX = 2147483647;
    // Function to check whether array A
    // can be made such that all operator
    // of array B holds true for each index i
    const checkComparisons = (N, A, B) => {
        // Declaring initial ranges as 0 to Infinity
        let left = 0, right = INT_MAX, index = 0;
        let valid = 1;
        for (let i = 0; i < N - 1; i++, index++) {
 
            // If current element is not empty
            if (A[i] != -1) {
 
                // Compare normal elements
                // (non empty i.e. != -1) based on B
                if (A[i + 1] != -1) {
                    if (B[index] == '>') {
                        if (A[i] <= A[i + 1])
                            return "No";
                    }
                    if (B[index] == '=') {
                        if (A[i] != A[i + 1])
                            return "No";
                    }
                    if (B[index] == '<') {
                        if (A[i] >= A[i + 1])
                            return "No";
                    }
                }
 
                // Compare and update ranges
                // for empty spaces
                else {
                    if (B[index] == '>') {
                        left = 0;
                        right = A[i] - 1;
                    }
                    if (B[index] == '=') {
                        left = A[i];
                        right = A[i];
                    }
                    if (B[index] == '<') {
                        left = A[i] + 1;
                        right = INT_MAX;
                    }
                }
            }
 
            // If current element is empty
            else {
 
                // Compare and update ranges if
                // next element is not empty
                if (A[i + 1] != -1) {
                    if (B[index] == '>') {
                        if (right <= A[i + 1])
                            return "No";
                    }
                    if (B[index] == '=') {
                        if ((left > A[i + 1])
                            || (right < A[i + 1]))
                            return "No";
                    }
                    if (B[index] == '<') {
                        if (left >= A[i + 1])
                            return "No";
                    }
                }
 
                // Compare and update ranges if
                // next element is empty based on
                // previous ranges
                else {
                    if (B[index] == '>') {
                        left = 0;
                        right--;
                    }
                    if (B[index] == '<') {
                        left++;
                        right = INT_MAX;
                    }
                }
            }
            if (left > right)
                return "No";
            if (right < 0)
                return "No";
            if (left < 0)
                return "No";
        }
        return "Yes";
    }
 
    // Driver Code
 
    let N = 6;
    let A = [8, 6, -1, -1, -1, 11];
    let B = ['>', '>', '=', '<', '>'];
 
    // Function call
    document.write(checkComparisons(N, A, B));
 
    // This code is contributed by rakeshsahni
 
</script>


Output

Yes

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


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