
Problem Statement
You are provided with two non-increasing integer arrays, nums1
and nums2
, both using 0-based indexing. A valid pair of indices (i, j)
is defined under two conditions: firstly, the index i
from nums1
must not exceed the index j
from nums2
(i <= j
), and secondly, the value at nums1[i]
must be less than or equal to the value at nums2[j]
(nums1[i] <= nums2[j]
). The distance of such a pair is the difference j - i
. Your task is to determine the maximum possible distance for such valid index pairs and return this value. If no valid pairs exist, the function should return 0. Each array element must follow the non-increasing order, which means each subsequent element must be less than or equal to its predecessor.
Examples
Example 1
Input:
nums1 = [55,30,5,4,2], nums2 = [100,20,10,10,5]
Output:
2
Explanation:
The valid pairs are (0,0), (2,2), (2,3), (2,4), (3,3), (3,4), and (4,4). The maximum distance is 2 with pair (2,4).
Example 2
Input:
nums1 = [2,2,2], nums2 = [10,10,1]
Output:
1
Explanation:
The valid pairs are (0,0), (0,1), and (1,1). The maximum distance is 1 with pair (0,1).
Example 3
Input:
nums1 = [30,29,19,5], nums2 = [25,25,25,25,25]
Output:
2
Explanation:
The valid pairs are (2,2), (2,3), (2,4), (3,3), and (3,4). The maximum distance is 2 with pair (2,4).
Constraints
1 <= nums1.length, nums2.length <= 105
1 <= nums1[i], nums2[j] <= 105
- Both
nums1
andnums2
are non-increasing.
Approach and Intuition
Given the constraints and examples, the brute-force approach of comparing all possible pairs (i, j)
with i
from nums1
and j
from nums2
is suboptimal due to potential high computational cost when array lengths are maximized. The following approach can be used for a more efficient solution:
Utilize a two-pointer technique to leverage the sorted (non-increasing) nature of
nums1
andnums2
.Set initial pointers at the start of both arrays. Begin with
i=0
fornums1
andj=0
fornums2
.Increase
j
untilnums1[i]
is less than or equal tonums2[j]
orj
reaches the end. This ensures compliance with the conditionnums1[i] <= nums2[j]
.If a valid
j
is found fornums1[i]
, calculatej - i
and update the maximum distance if this value exceeds the current known maximum.Move to the next element in
nums1
by incrementingi
and try to find a newj
starting from the currentj
(sincenums2
is non-increasing, a valid j fornums1[i]
will be valid for all subsequenti
).Repeat this until you have traversed all elements of
nums1
while ensuring the above conditions.
- Return the recorded maximum distance. If no valid pairs were found (if the maximum distance wasn't updated), return 0.
This method works efficiently as each pointer only moves forward through its respective array, ensuring that every element is considered exactly once, achieving optimal performance in respect to time complexity.
Solutions
- C++
- Java
- Python
class Solution {
public:
int findMaxDistance(vector<int>& array1, vector<int>& array2) {
int maxDist = 0, index1 = 0, index2 = 0;
while (index1 < array1.size() && index2 < array2.size()) {
if (array1[index1] > array2[index2]) {
index1++;
} else {
maxDist = max(maxDist, index2 - index1);
index2++;
}
}
return maxDist;
}
};
This C++ solution addresses the problem of finding the maximum distance between a pair of values from two vectors where the value from the second vector is not smaller than the value from the first. The method findMaxDistance
performs this check and determines the maximum permissible distance.
- The function initializes
maxDist
,index1
, andindex2
to zero to keep track of the maximum distance and the current indices inarray1
andarray2
. - It employs a while loop to iterate through both arrays as long as neither index exceeds its respective array's length. This ensures all possible pairs of elements are considered.
- Inside the loop, a condition checks if the current element in
array1
is greater than the element inarray2
at their respective indices. If true, it incrementsindex1
to check the next element inarray1
. - If the element in
array1
is not greater, it calculates the distance between the indices (index2 - index1
) and updatesmaxDist
if this new distance is greater than the previously recorded maximum distance. Subsequently,index2
is incremented. - After the loop concludes,
maxDist
holds the required maximum distance, which is then returned.
This approach ensures the solution is optimized and only viable pairs (where array1[index]
is not greater than array2[index]
) are considered, improving both clarity and efficiency of the algorithm.
class Solution {
public int maximumSeparation(int[] array1, int[] array2) {
int maxDiff = 0, index1 = 0, index2 = 0;
while (index1 < array1.length && index2 < array2.length) {
if (array1[index1] > array2[index2]) {
index1++;
} else {
maxDiff = Math.max(maxDiff, index2 - index1);
index2++;
}
}
return maxDiff;
}
}
This Java program aims to find the maximum distance between elements from two arrays under specific comparison conditions. The function maximumSeparation
accepts two integer arrays, array1
and array2
, and uses two pointers, index1
and index2
, to traverse these arrays.
- Initialize
maxDiff
to track the maximum separation distance. - Use a
while
loop to iterate through both arrays till one of them is fully traversed. - Compare elements from
array1
andarray2
:- If
array1[index1]
is greater thanarray2[index2]
, incrementindex1
. - Otherwise, calculate the difference between the current positions (
index2
-index1
) to updatemaxDiff
, and incrementindex2
.
- If
The function returns the value of maxDiff
, which represents the largest valid index difference for which the value at array1[index1]
is not greater than array2[index2]
. This approach ensures an efficient O(n+m) complexity where n and m are the lengths of array1
and array2
, respectively, by using a linear pass through the arrays without the need for nested loops.
class Solution:
def maximumGap(self, arr1: List[int], arr2: List[int]) -> int:
max_gap = 0
idx1, idx2 = 0, 0
while idx1 < len(arr1) and idx2 < len(arr2):
if arr1[idx1] > arr2[idx2]:
idx1 += 1
else:
max_gap = max(max_gap, idx2 - idx1)
idx2 += 1
return max_gap
The provided Python script defines a method to calculate the maximum gap in terms of indices between two sorted arrays, arr1
and arr2
. The arrays must contain integers. The function, maximumGap
, iterates through both arrays by using two pointers (idx1
for arr1
and idx2
for arr2
) and computes the maximum gap where the value in arr1
is not greater than the value in arr2
.
Follow these steps to understand the process implemented:
- Initialize
max_gap
to 0, and pointer indicesidx1
andidx2
to 0. - Use a
while
loop to traverse elements in both arrays until one of the pointers reaches the end of its respective array. - Compare the current elements in
arr1
andarr2
using the pointers. - If the element in
arr1
is greater, incrementidx1
. - If the element in
arr1
is less than or equal to the element inarr2
, updatemax_gap
if the current gap (idx2 - idx1
) is larger than the recordedmax_gap
, then incrementidx2
. - Continue to loop until the end of one of the arrays.
- Return
max_gap
, which is the maximum distance in indices between corresponding elements in the two arrays that meet the condition (arr1[idx1]
<=arr2[idx2]
).
This method is efficient with array traversal due to the use of two-pointer technique. Ensure that both arrays are sorted to properly utilize this method.
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