Learn and practice the Two Pointers, and Modified Binary Search patterns to solve coding interview problems.

Speedrun the Google Coding Interview: Top Problems in Java

This module groups together two of the most common patterns that may be used to solve problems where the input data is linear: Two Pointers and Modified Binary Search. Both the Two Pointers and the Modified Binary Search patterns derive their power from their ability to efficiently reduce the size of the search space, that is, both exemplify the divide and conquer approach to problem-solving. After completing this module, we will have the knowledge of how to use this set of related patterns to solve a diverse range of problems.

Two Pointers & Co.

# Statement

You are given a sorted integer array, `nums`, and an integer, `target`. The array may have been rotated by an arbitrary number. Your task is to find and return the index of `target` in this array. If `target` does not exist, return -1.

An original sorted array before rotation is given below:

digraph g {
 rankdir = "TB" 
 splines = false;
 bgcolor = transparent
 node[shape = record, fontname = "courier", fontsize="18"];
 size = "9"
 bgcolor = "transparent"
 array[label = <<table border="0" cellborder="1" cellspacing="0"> \
 
 <tr> \
 <td height="35" width="45" port="e1" bgcolor="#FFDC5C" >1</td>\ 
 <td height="35" width="45" port="e0" bgcolor="#FFDC5C" >10</td>\
 <td height="35" width="45" port="e2" bgcolor="#FFDC5C" >20</td> \ 
 <td height="35" width="45" port="e3" bgcolor="#FFDC5C" >47</td> \ 
 <td height="35" width="45" port="e4" bgcolor="#FFDC5C" >59</td> \ 
 <td height="35" width="45" port="e5" bgcolor="#FFDC5C" >63</td> \ 
 <td height="35" width="45" port="e6" bgcolor="#FFDC5C" >75</td> \ 
 <td height="35" width="45" port="e7" bgcolor="#FFDC5C" >88</td> \ 
 <td height="35" width="45" port="e8" bgcolor="#FFDC5C" >99</td> \ 
 <td height="35" width="45" port="e9" bgcolor="#FFDC5C" >107</td> \
 <td height="35" width="45" port="e10" bgcolor="#FFDC5C">120</td> \
 <td height="35" width="45" port="e11" bgcolor="#FFDC5C">133</td> \
 <td height="35" width="45" port="e12" bgcolor="#FFDC5C">155</td> \
 <td height="35" width="45" port="e13" bgcolor="#FFDC5C">162</td> \
 <td height="35" width="45" port="e14" bgcolor="#FFDC5C">176</td> \
 <td height="35" width="45" port="e15" bgcolor="#FFDC5C">188</td> \
 <td height="35" width="45" port="e16" bgcolor="#FFDC5C">199</td> \
 <td height="35" width="45" port="e17" bgcolor="#FFDC5C">200</td> \
 <td height="35" width="45" port="e18" bgcolor="#FFDC5C">210</td> \
 <td height="35" width="45" port="e19" bgcolor="#FFDC5C">222</td> \
 </tr> \
 </table>> shape=none];
 
 }

After rotating this array 6 times, it changes to:

digraph g {
 rankdir = "TB" 
 splines = false;
 bgcolor = transparent
 node[shape = record, fontname = "courier", fontsize="18"];
 size = "9"
 bgcolor = "transparent"
 array[label = <<table border="0" cellborder="1" cellspacing="0"> \
 
 <tr> \
 						<td height="35" width="45" port="e0" bgcolor="#33CFFF">176</td>\
 <td height="35" width="45" port="e0" bgcolor="#33CFFF" >188</td>\
 <td height="35" width="45" port="e2" bgcolor="#33CFFF" >199</td> \ 
 <td height="35" width="45" port="e3" bgcolor="#33CFFF" >200</td> \ 
 <td height="35" width="45" port="e4" bgcolor="#33CFFF" >210</td> \ 
 <td height="35" width="45" port="e5" bgcolor="#33CFFF" >222</td> \ 
 <td height="35" width="45" port="e6" bgcolor="#FFDC5C" >1</td> \ 
 <td height="35" width="45" port="e7" bgcolor="#FFDC5C" >10</td> \ 
 <td height="35" width="45" port="e8" bgcolor="#FFDC5C" >20</td> \ 
 <td height="35" width="45" port="e9" bgcolor="#FFDC5C" >47</td> \
 <td height="35" width="45" port="e10" bgcolor="#FFDC5C">59</td> \
 <td height="35" width="45" port="e11" bgcolor="#FFDC5C">63</td> \
 <td height="35" width="45" port="e12" bgcolor="#FFDC5C">75</td> \
 <td height="35" width="45" port="e13" bgcolor="#FFDC5C">88</td> \
 <td height="35" width="45" port="e14" bgcolor="#FFDC5C">99</td> \
 <td height="35" width="45" port="e15" bgcolor="#FFDC5C">107</td> \
 <td height="35" width="45" port="e16" bgcolor="#FFDC5C">120</td> \
 <td height="35" width="45" port="e17" bgcolor="#FFDC5C">133</td> \
 <td height="35" width="45" port="e18" bgcolor="#FFDC5C">155</td> \
 <td height="35" width="45" port="e18" bgcolor="#FFDC5C">162</td> \
 </tr> \
 </table>> shape=none];
 
 }

**Constraints**

- All values in `nums` are unique.
- The values in `nums` are sorted in ascending order.
- The array may have been rotated by some arbitrary number.
- $1 \leq$ `nums.length` $\leq 1000$
- $-10^4 \leq$ `nums[i]` $\leq 10^4$
- $-10^4 \leq$ `target` $\leq 10^4$

# Examples

What is the output if the following rotated sorted array and target are given as input?

nums = [4, 5, 6, 7, 0, 1, 2]

target = 1



What is the output if the following rotated sorted array and target are given as input?
nums = [4, 5, 6, 7, 0, 1, 2]
target = 1

What is the output if the following rotated sorted array and target are given as input?

nums = [7, 8, 9, 0, 6]

target = 1

What is the output if the following rotated sorted array and target are given as input?
nums = [7, 8, 9, 0, 6]
target = 1

What is the output if the following rotated sorted array and target are given as input?

nums = [15, 16, 19, 20, 25, 1, 3, 4, 5, 7, 10, 14]

target = 25

What is the output if the following rotated sorted array and target are given as input?
nums = [15, 16, 19, 20, 25, 1, 3, 4, 5, 7, 10, 14]
target = 25

Search in Rotated Sorted Array

Search in Rotated Sorted Array

import java.util.*;

public class Solution {
   public static int binarySearchRotated(int[] nums, int target) {
      
      // Replace this placeholder return statement with your code
      return -1;
   }
}

# Quick Fix\nYou are Edward, a Senior Software Engineer who specializes in coding interview patterns, data structures and algorithms, and is great at debugging issues in code.\n\n- Edward, please keep in mind that you do not have to print the data of the placeholders. Also, do not mention anything about yourself.\n\nEdward, you are given the following data in the placeholders:\n- Function name: `binarySearchRotated`\n- Candidate's submitted solution code: USER_CODE\n- Error: ERROR_DESCRIPTION\n- Failed test case: FAILED_TEST_CASE\n- Coding language: `Java`\n\n- Edward, your task is to understand the type of error that is occurring in the coding language provided.\n- Once you've understood the error as per the error’s description, and why the test case failed, you will suggest how the candidate can improve their solution code to fix the error.\n- Edward, you will not give the corrected code, just a hint on what the error tells, and how to fix it.\n- Edward, please limit your response strictly to 100 words. No more words than the limit.

# Review Code\nEdward, your current task is to review the `binarySearchRotated` function(s) and the other function(s) invoked in it/them.\n\n-Take your time and review the candidate's solution code.\n- Do not review any extra/helper files or functions not invoked in the solution function `binarySearchRotated`.\n- Do not review the code of data structure objects instantiated like a binary tree, linked list, trie, union find, etc., in the solution functions. Just review the solution/algorithm function of the problem. \n- Also, if the solution code doesn’t use the coding pattern stated above, then please encourage the candidate to reattempt the problem using that pattern corectly.

# Review time complexity\n- Please provide a brief time complexity analysis of the function(s) `binarySearchRotated`. \n- While calculating time complexity, include the time complexity of built-in Java functions invoked by the code. \n- The answer should be to the point in approximately 60 words.

# Review space complexity\n- Please provide a brief space complexity analysis of the function(s) `binarySearchRotated`. \n- While calculating space complexity, include the space complexity of built-in Java functions invoked by the code. \n- Do not include the space used for the input parameters and the space used for just returning an output (not used as a scratch space). \n- The answer should be to the point in approximately 60 words.

OneDArray

import ds_v1.DataStructure.DataStructure;
import java.util.Arrays;
import java.io.IOException;
import com.fasterxml.jackson.databind.ObjectMapper;
public class OneDArray extends DataStructure {
    int[] values = {};


    public OneDArray() {
        //values = {};
    }
    

    public OneDArray(int[] val){
        this.values = val.clone();
        
    }

    public String toString() {
        return Arrays.toString(values);
    }

    @Override
    public OneDArray construct(String new_value, Integer type_of_ds) {
        int[] array={};
        try {
            ObjectMapper mapper = new ObjectMapper();
            array = mapper.readValue(new_value, int[].class);
            
        } catch (IOException e) {
            e.printStackTrace();
        }
        return new OneDArray(array);

    }

    @Override
    public String traverse(Object input, Integer type_of_ds) {
        
        return input.toString();
    }

}

def evaluate_results(self) -> object:
 """
 self.__inputs is the list of inputs that are passed as input to function
 self.__actual_output is a list of outputs that is produced by the learner's code
 return : self.__edu__result object
 """

 def binary_search(nums, low, high, target):

 if (low > high):
 return -1
 mid = low + (high - low) // 2

 if nums[mid] == target:
 return mid

 if nums[low] <= nums[mid]:
 if nums[low] <= target and target < nums[mid]:
 return binary_search(nums, low, mid-1, target)
 return binary_search(nums, mid+1, high, target)
 else:
 if nums[mid] < target and target <= nums[high]:
 return binary_search(nums, mid+1, high, target)
 return binary_search(nums, low, mid-1, target)


 def binary_search_rotated(nums, target):
 return binary_search(nums, 0, len(nums)-1, target)

 expected_outputs = [] 
 expected = binary_search_rotated(self.__inputs[0], self.__inputs[1])
 actual_output = self.__actual_outputs[0]
 

 if expected == actual_output:
 expected_outputs.append(expected)
 self.__edu_result.store_result(expected_outputs, TestCaseResult.PASS)
 else:
 expected_outputs.append(expected)
 self.__edu_result.store_result(expected_outputs, TestCaseResult.FAIL)

 return self.__edu_result

import java.util.*;

public class Main {
   public static int driver(OneDArray nums, int target) {
      
      // Replace this placeholder return statement with your code
      return Solution.binarySearchRotated(nums.values, target);
   }
}


search_in_rotated_sorted_array.yaml

testcases:
  - name: "testcase 1"
    inputs:
    - 1 : [6, 7, 1, 2, 3, 4, 5]
    - 2 : 3
    output: 
    - 1 : 4
  - name: "testcase 2"
    inputs:
    - 1: [6, 7, 1, 2, 3, 4, 5]
    - 2: 6
    output:
    - 1 : 0
  - name: "testcase 3"
    inputs:
    - 1: [4, 5, 6, 1, 2, 3]
    - 2: 3
    output:
    - 1 : 5
  - name: "testcase 4"
    inputs:
    - 1: [4, 5, 6, 1, 2, 3]
    - 2: 6
    output:
    - 1 : 2
  - name: "testcase 5"
    inputs:
    - 1: [4]
    - 2: 1
    output:
    - 1 : -1



def sum(input_1, input_2):
    return input_1+input_2

def validate_inputs(self) -> None:
 """
 self.__inputs is the list of inputs that are passed as input to function
 store the result using the self.store_result(error_msg, status)
 return : void
 """

 def is_valid_1d_array(arr):
 if not isinstance(arr, list):
 return False

 return all(isinstance(elem, int) and not isinstance(elem, bool) for elem in arr)


 def is_sorted(arr):
 for i in range(len(arr) - 1):
 if arr[i] > arr[i + 1]:
 return False
 return True

 def is_sorted_rotated(arr):
 if len(arr) == 1 or is_sorted(arr) == True:
 return True
 
 right = left = []
 for i in range(len(arr)):
 if i + 1 < len(arr):
 if arr[i] > arr[i + 1]:
 left = arr[:i + 1]
 right = arr[i + 1:]

 if is_sorted(left) and is_sorted(right) and left[0] > right[len(right)-1]:
 return True
 else:
 return False

 nums = self.__inputs[0]
 target = self.__inputs[1]

 if not is_valid_1d_array(nums):
 self.store_result ("'nums' should be a valid 1D array of integers", ValidationFunction.FAIL)
 elif not 1 <= len(nums) <= 1000:
 self.store_result ("'nums' length should be in the range [1, 10000]", ValidationFunction.FAIL)
 elif not -10000 <= target <= 10000:
 self.store_result ("'target' should be in the range [-10000, 10000]", ValidationFunction.FAIL)
 elif len(nums) != len(list(set(nums))):
 self.store_result ("'nums' should have unique elements", ValidationFunction.FAIL)
 elif is_sorted_rotated(nums) == False:
 self.store_result ("'nums' should be rotated and sorted in ascending order", ValidationFunction.FAIL)
 elif any(not -10000 <= i <= 10000 for i in nums):
 self.store_result ("'nums[i]' should be in the range [-10000, 10000]", ValidationFunction.FAIL)
 else:
 self.store_result("", ValidationFunction.PASS)

Try to solve the Search in Rotated Sorted Array problem.

Two Pointers

Modified Binary Search

Conclusion

Search in Rotated Sorted Array

Statement

Examples

Understand the problem

Figure it out!

Try it yourself