Solutions

Prerequisites

Before attempting this problem, you should be comfortable with:

Hash Maps - Used to store value-to-index mappings for O(1) lookup
Array Traversal - Basic iteration to build mappings between two arrays
Bit Manipulation (Optional) - The advanced solution encodes indices into values using bit shifts

1. Brute Force

Intuition

The problem asks us to find, for each element in nums1, an index in nums2 where that same value appears. The simplest approach is to check every position in nums2 for each element in nums1. When we find a match, we record that index and move on. This guarantees correctness since we examine all possibilities.

Algorithm

Create a result array mappings of the same length as nums1.
For each index i in nums1:
- Iterate through nums2 with index j.
- When nums1[i] equals nums2[j], store j in mappings[i] and break.
Return the mappings array.

class Solution:
    def anagramMappings(self, nums1: List[int], nums2: List[int]) -> List[int]:
        # List to store the anagram mappings.
        mappings = [0] * len(nums1)
        
        for i in range(len(nums1)):
            for j in range(len(nums2)):
                # Store the corresponding index of number in the second list.
                if nums1[i] == nums2[j]:
                    mappings[i] = j
                    break
        
        return mappings

class Solution {
    public int[] anagramMappings(int[] nums1, int[] nums2) {
        // List to store the anagram mappings.
        int[] mappings = new int[nums1.length];
        
        for (int i = 0; i < nums1.length; i++) {
            for (int j = 0; j < nums2.length; j++) {
                // Store the corresponding index of number in the second list.
                if (nums1[i] == nums2[j]) {
                    mappings[i] = j;
                    break;
                }
            }
        }
        return mappings;
    }
}

class Solution {
public:
    vector<int> anagramMappings(vector<int>& nums1, vector<int>& nums2) {
        // List to store the anagram mappings.
        vector<int> mappings;
        
        for (int num : nums1) {
            for (int i = 0; i < nums2.size(); i++) {
                // Store the corresponding index of number in the second list.
                if (num == nums2[i]) {
                    mappings.push_back(i);
                    break;
                }
            }
        }
        return mappings;
    }
};

class Solution {
    /**
     * @param {number[]} nums1
     * @param {number[]} nums2
     * @return {number[]}
     */
    anagramMappings(nums1, nums2) {
        // Array to store the anagram mappings.
        const mappings = new Array(nums1.length);

        for (let i = 0; i < nums1.length; i++) {
            for (let j = 0; j < nums2.length; j++) {
                // Store the corresponding index of number in the second array.
                if (nums1[i] === nums2[j]) {
                    mappings[i] = j;
                    break;
                }
            }
        }
        return mappings;
    }
}

func anagramMappings(nums1 []int, nums2 []int) []int {
    // Slice to store the anagram mappings.
    mappings := make([]int, len(nums1))

    for i := 0; i < len(nums1); i++ {
        for j := 0; j < len(nums2); j++ {
            // Store the corresponding index of number in the second slice.
            if nums1[i] == nums2[j] {
                mappings[i] = j
                break
            }
        }
    }
    return mappings
}

class Solution {
    fun anagramMappings(nums1: IntArray, nums2: IntArray): IntArray {
        // Array to store the anagram mappings.
        val mappings = IntArray(nums1.size)

        for (i in nums1.indices) {
            for (j in nums2.indices) {
                // Store the corresponding index of number in the second array.
                if (nums1[i] == nums2[j]) {
                    mappings[i] = j
                    break
                }
            }
        }
        return mappings
    }
}

class Solution {
    func anagramMappings(_ nums1: [Int], _ nums2: [Int]) -> [Int] {
        // Array to store the anagram mappings.
        var mappings = [Int](repeating: 0, count: nums1.count)

        for i in 0..<nums1.count {
            for j in 0..<nums2.count {
                // Store the corresponding index of number in the second array.
                if nums1[i] == nums2[j] {
                    mappings[i] = j
                    break
                }
            }
        }
        return mappings
    }
}

Time & Space Complexity

Time complexity: $O(N^2)$
Space complexity: $O(1)$ constant space

Where $N$ is the number of integers in the list nums1 and nums2.

2. HashMap

Intuition

Instead of scanning nums2 repeatedly, we can preprocess it into a hash map that stores each value's index. This allows us to look up the corresponding index for any element in constant time. Since both arrays are anagrams, every element in nums1 is guaranteed to exist in nums2.

Algorithm

Build a hash map valueToPos where each key is a value from nums2 and each entry stores its index.
Create a result array mappings.
For each element in nums1, look up its index in the hash map and store it in mappings.
Return mappings.

class Solution:
    def anagramMappings(self, nums1: List[int], nums2: List[int]) -> List[int]:
        # Store the index corresponding to the value in the second list.
        valueToPos = {}
        for i in range(len(nums2)):
            valueToPos[nums2[i]] = i

        # List to store the anagram mappings.
        mappings = [0] * len(nums1)
        for i in range(len(nums1)):
            mappings[i] = valueToPos[nums1[i]]

        return mappings

class Solution {
    public int[] anagramMappings(int[] nums1, int[] nums2) {
        // Store the index corresponding to the value in the second list.
        HashMap<Integer,Integer> valueToPos = new HashMap<>();
        for (int i = 0; i < nums2.length; i++) {
            valueToPos.put(nums2[i], i);
        }

        // List to store the anagram mappings.
        int[] mappings = new int[nums1.length];
        for (int i = 0; i < nums1.length; i++) {
            mappings[i]  = valueToPos.get(nums1[i]);
        }

        return mappings;
    }
}

class Solution {
public:
    vector<int> anagramMappings(vector<int>& nums1, vector<int>& nums2) {
        // Store the index corresponding to the value in the second list.
        unordered_map<int, int> valueToPos;
        for (int i = 0; i < nums2.size(); i++) {
            valueToPos[nums2[i]] = i;
        }
        
        // List to store the anagram mappings.
        vector<int> mappings;
        for (int num : nums1) {
            mappings.push_back(valueToPos[num]);
        }
        
        return mappings;
    }
};

class Solution {
    /**
     * @param {number[]} nums1
     * @param {number[]} nums2
     * @return {number[]}
     */
    anagramMappings(nums1, nums2) {
        // Store the index corresponding to the value in the second list.
        const valueToPos = new Map();
        for (let i = 0; i < nums2.length; i++) {
            valueToPos.set(nums2[i], i);
        }

        // List to store the anagram mappings.
        const mappings = new Array(nums1.length);
        for (let i = 0; i < nums1.length; i++) {
            mappings[i] = valueToPos.get(nums1[i]);
        }

        return mappings;
    }
}

func anagramMappings(nums1 []int, nums2 []int) []int {
    // Store the index corresponding to the value in the second slice.
    valueToPos := make(map[int]int)
    for i := 0; i < len(nums2); i++ {
        valueToPos[nums2[i]] = i
    }

    // Slice to store the anagram mappings.
    mappings := make([]int, len(nums1))
    for i := 0; i < len(nums1); i++ {
        mappings[i] = valueToPos[nums1[i]]
    }

    return mappings
}

class Solution {
    fun anagramMappings(nums1: IntArray, nums2: IntArray): IntArray {
        // Store the index corresponding to the value in the second array.
        val valueToPos = HashMap<Int, Int>()
        for (i in nums2.indices) {
            valueToPos[nums2[i]] = i
        }

        // Array to store the anagram mappings.
        val mappings = IntArray(nums1.size)
        for (i in nums1.indices) {
            mappings[i] = valueToPos[nums1[i]]!!
        }

        return mappings
    }
}

class Solution {
    func anagramMappings(_ nums1: [Int], _ nums2: [Int]) -> [Int] {
        // Store the index corresponding to the value in the second array.
        var valueToPos = [Int: Int]()
        for i in 0..<nums2.count {
            valueToPos[nums2[i]] = i
        }

        // Array to store the anagram mappings.
        var mappings = [Int](repeating: 0, count: nums1.count)
        for i in 0..<nums1.count {
            mappings[i] = valueToPos[nums1[i]]!
        }

        return mappings
    }
}

Time & Space Complexity

Time complexity: $O(N)$
Space complexity: $O(N)$

Where $N$ is the number of integers in the list nums1 and nums2.

3. Bit Manipulation + Sorting

Intuition

We can avoid using extra hash map space by encoding each element's original index directly into the element itself using bit manipulation. By left-shifting each value and adding its index, we preserve both pieces of information in a single integer. After sorting both arrays, elements with the same original value will align at the same positions, letting us extract and pair up their indices.

Algorithm

For each index i, encode both arrays: nums[i] = (nums[i] << 7) + i. The shift amount (7 bits) must be large enough to hold the maximum index.
Sort both nums1 and nums2. Equal values now appear at matching positions.
Create the result array mappings.
For each position i, extract the original indices using a bitmask: mappings[nums1[i] & mask] = nums2[i] & mask.
Return mappings.

class Solution:
    def anagramMappings(self, nums1: List[int], nums2: List[int]) -> List[int]:
        bitsToShift = 7
        numToGetLastBits = (1 << bitsToShift) - 1
        
        # Store the index within the integer itself.
        for i in range(len(nums1)):
            nums1[i] = (nums1[i] << bitsToShift) + i
            nums2[i] = (nums2[i] << bitsToShift) + i
        
        # Sort both lists so that the original integers end up at the same index.
        nums1.sort()
        nums2.sort()
        
        # List to store the anagram mappings.
        mappings = [0] * len(nums1)
        
        for i in range(len(nums1)):
            # Store the index in the second list corresponding to the integer index in the first list.
            mappings[nums1[i] & numToGetLastBits] = (nums2[i] & numToGetLastBits)
        
        return mappings

class Solution {
    final int bitsToShift = 7;
    final int numToGetLastBits = (1 << bitsToShift) - 1;

    public int[] anagramMappings(int[] nums1, int[] nums2) {
        // Store the index within the integer itself.
        for (int i = 0; i < nums1.length; i++) {
            nums1[i] = (nums1[i] << bitsToShift) + i;
            nums2[i] = (nums2[i] << bitsToShift) + i;
        }

        // Sort both lists so that the original integers end up at the same index.
        Arrays.sort(nums1);
        Arrays.sort(nums2);

        // List to store the anagram mappings.
        int[] mappings = new int[nums1.length];
        for (int i = 0; i < nums1.length; i++) {
            // Store the index in the second list corresponding to the integer index in the first list.
            mappings[nums1[i] & numToGetLastBits] = (nums2[i] & numToGetLastBits);
        }

        return mappings;
    }
}

class Solution {
public:
    const int bitsToShift = 7;
    const int numToGetLastBits = (1 << bitsToShift) - 1;
    
    vector<int> anagramMappings(vector<int>& nums1, vector<int>& nums2) {
        // Store the index within the integer itself.
        for (int i = 0; i < nums1.size(); i++) {
            nums1[i] = (nums1[i] << bitsToShift) + i;
            nums2[i] = (nums2[i] << bitsToShift) + i;
        }
        
        // Sort both lists so that the original integers end up at the same index.
        sort(nums1.begin(), nums1.end());
        sort(nums2.begin(), nums2.end());
        
        // List to store the anagram mappings.
        vector<int> mappings(nums1.size());
        for (int i = 0; i < nums1.size(); i++) {
            // Store the index in the second list corresponding to the integer index in the first list.
            mappings[nums1[i] & numToGetLastBits] = (nums2[i] & numToGetLastBits);
        }

        return mappings;
    }
};

class Solution {
    /**
     * @param {number[]} nums1
     * @param {number[]} nums2
     * @return {number[]}
     */
    anagramMappings(nums1, nums2) {
        const bitsToShift = 7;
        const numToGetLastBits = (1 << bitsToShift) - 1;

        // Store the index within the integer itself.
        for (let i = 0; i < nums1.length; i++) {
            nums1[i] = (nums1[i] << bitsToShift) + i;
            nums2[i] = (nums2[i] << bitsToShift) + i;
        }

        // Sort both arrays so that the original integers end up at the same index.
        nums1.sort((a, b) => a - b);
        nums2.sort((a, b) => a - b);

        // Array to store the anagram mappings.
        const mappings = new Array(nums1.length);

        for (let i = 0; i < nums1.length; i++) {
            // Store the index in the second array corresponding to the integer index in the first array.
            mappings[nums1[i] & numToGetLastBits] = (nums2[i] & numToGetLastBits);
        }

        return mappings;
    }
}

func anagramMappings(nums1 []int, nums2 []int) []int {
    bitsToShift := 7
    numToGetLastBits := (1 << bitsToShift) - 1

    // Store the index within the integer itself.
    for i := 0; i < len(nums1); i++ {
        nums1[i] = (nums1[i] << bitsToShift) + i
        nums2[i] = (nums2[i] << bitsToShift) + i
    }

    // Sort both slices so that the original integers end up at the same index.
    sort.Ints(nums1)
    sort.Ints(nums2)

    // Slice to store the anagram mappings.
    mappings := make([]int, len(nums1))

    for i := 0; i < len(nums1); i++ {
        // Store the index in the second slice corresponding to the integer index in the first slice.
        mappings[nums1[i] & numToGetLastBits] = nums2[i] & numToGetLastBits
    }

    return mappings
}

class Solution {
    fun anagramMappings(nums1: IntArray, nums2: IntArray): IntArray {
        val bitsToShift = 7
        val numToGetLastBits = (1 shl bitsToShift) - 1

        // Store the index within the integer itself.
        for (i in nums1.indices) {
            nums1[i] = (nums1[i] shl bitsToShift) + i
            nums2[i] = (nums2[i] shl bitsToShift) + i
        }

        // Sort both arrays so that the original integers end up at the same index.
        nums1.sort()
        nums2.sort()

        // Array to store the anagram mappings.
        val mappings = IntArray(nums1.size)

        for (i in nums1.indices) {
            // Store the index in the second array corresponding to the integer index in the first array.
            mappings[nums1[i] and numToGetLastBits] = nums2[i] and numToGetLastBits
        }

        return mappings
    }
}

class Solution {
    func anagramMappings(_ nums1: [Int], _ nums2: [Int]) -> [Int] {
        var nums1 = nums1
        var nums2 = nums2
        let bitsToShift = 7
        let numToGetLastBits = (1 << bitsToShift) - 1

        // Store the index within the integer itself.
        for i in 0..<nums1.count {
            nums1[i] = (nums1[i] << bitsToShift) + i
            nums2[i] = (nums2[i] << bitsToShift) + i
        }

        // Sort both arrays so that the original integers end up at the same index.
        nums1.sort()
        nums2.sort()

        // Array to store the anagram mappings.
        var mappings = [Int](repeating: 0, count: nums1.count)

        for i in 0..<nums1.count {
            // Store the index in the second array corresponding to the integer index in the first array.
            mappings[nums1[i] & numToGetLastBits] = nums2[i] & numToGetLastBits
        }

        return mappings
    }
}

Time & Space Complexity

Time complexity: $O(N \log N)$
Space complexity: $O(\log N)$

Where $N$ is the number of integers in the list nums1 and nums2.

Common Pitfalls

Overwriting Indices for Duplicate Values

When using a hash map, storing only a single index per value means duplicate values all map to the same index. If the problem requires distinct mappings for duplicates, use a list of indices or pop from a stack of indices instead.

Off-by-One Errors in Bit Manipulation

When encoding indices into values using bit shifts, using too few bits causes index collisions for larger arrays. Ensure the shift amount is large enough to accommodate the maximum possible index value.