Given two arrays that are duplicates of each other except one element, that is one element from one of the array is missing, we need to find that missing element.
Examples:
Input: arr1[] = {1, 4, 5, 7, 9}
arr2[] = {4, 5, 7, 9}
Output: 1
1 is missing from second array.
Input: arr1[] = {2, 3, 4, 5}
arr2[] = {2, 3, 4, 5, 6}
Output: 6
6 is missing from first array.
One simple solution is to iterate over arrays and check element by element and flag the missing element when an unmatched element is found, but this solution requires linear time oversize of the array.
Another efficient solution is based on a binary search approach. Algorithm steps are as follows:
- Start a binary search in a bigger array and get mid as (lo + hi) / 2
- If the value from both arrays is the same then the missing element must be in the right part so set lo as mid
- Else set hi as mid because the missing element must be in the left part of the bigger array if mid-elements are not equal.
- A special case is handled separately as for single element and zero elements array, the single element itself will be the missing element.
If the first element itself is not equal then that element will be the missing element./li>
Below is the implementation of the above steps
Javascript
<script> // Javascript program to find missing element from // same arrays (except one missing element) /* Function to find missing element based on binary search approach. arr1[] is of larger size and N is size of it.arr1[] and arr2[] are assumed to be in same order. */ function findMissingUtil(arr1, arr2, N) { // special case, for only element // which is missing in second array if (N == 1) return arr1[0]; // special case, for first // element missing if (arr1[0] != arr2[0]) return arr1[0]; // Initialize current corner points let lo = 0, hi = N - 1; // loop until lo < hi while (lo < hi) { let mid = parseInt((lo + hi) / 2, 10); // If element at mid indices are // equal then go to right subarray if (arr1[mid] == arr2[mid]) lo = mid; else hi = mid; // if lo, hi becomes // contiguous, break if (lo == hi - 1) break; } // missing element will be at hi // index of bigger array return arr1[hi]; } // This function mainly does basic error // checking and calls findMissingUtil function findMissing(arr1, arr2, M, N) { if (N == M - 1) document.write("Missing Element is " + findMissingUtil(arr1, arr2, M) + "</br>"); else if (M == N - 1) document.write("Missing Element is " + findMissingUtil(arr2, arr1, N) + "</br>"); else document.write("Invalid Input" + "</br>"); } let arr1 = [ 1, 4, 5, 7, 9 ]; let arr2 = [ 4, 5, 7, 9 ]; let M = arr1.length; let N = arr2.length; findMissing(arr1, arr2, M, N); </script> |
Output :
Missing Element is 1
Time Complexity: O(logM + logN), where M and N represents the size of the given two arrays.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
What if input arrays are not in the same order?
In this case, the missing element is simply XOR of all elements of both arrays. Thanks to Yolo Song for suggesting this.
Javascript
<script> // Javascript program to find // missing element from one array // such that it has all elements // of other array except one. // Elements in two arrays // can be in any order. // This function mainly does XOR // of all elements // of arr1[] and arr2[] function findMissing(arr1, arr2, M, N) { if (M != N-1 && N != M-1) { document.write("Invalid Input"); return; } // Do XOR of all element let res = 0; for (let i=0; i<M; i++) res = res^arr1[i]; for (let i=0; i<N; i++) res = res^arr2[i]; document.write("Missing element is " + res); } let arr1 = [4, 1, 5, 9, 7]; let arr2 = [7, 5, 9, 4]; let M = arr1.length; let N = arr2.length; findMissing(arr1, arr2, M, N); </script> |
Output :
Missing Element is 1
Time Complexity: O(M + N), where M and N represents the size of the given two arrays.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Please refer complete article on Find lost element from a duplicated array for more details!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
