How hard is LeetCode 1983. Widest Pair of Indices With Equal Range Sum?

LeetCode 1983. Widest Pair of Indices With Equal Range Sum is rated Medium on LeetCode.

What is the time complexity of LeetCode 1983. Widest Pair of Indices With Equal Range Sum?

The Python solution on this page runs in O(n).

What is the space complexity of LeetCode 1983. Widest Pair of Indices With Equal Range Sum?

The Python solution on this page uses O(n), where n is the length of the array nums auxiliary space.

What topics does LeetCode 1983. Widest Pair of Indices With Equal Range Sum cover?

LeetCode 1983. Widest Pair of Indices With Equal Range Sum is tagged Array, Hash Table and Prefix Sum on LeetCode.

Is LeetCode 1983. Widest Pair of Indices With Equal Range Sum a premium problem?

Yes. LeetCode 1983. Widest Pair of Indices With Equal Range Sum is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.

MediumNo. 1983LeetCode Premium

Widest Pair of Indices With Equal Range Sum — LeetCode 1983 Python Solution

Pattern: Prefix Sum
Topics: Array, Hash Table, Prefix Sum
Reading time: 2 min
Source: leetcode.com

The problem

You are given two 0-indexed binary arrays nums1 and nums2. Find the widest pair of indices (i, j) such that i <= j and nums1[i] + nums1[i+1] + ...

Example

Input: nums1 = [1,1,0,1], nums2 = [0,1,1,0]
Output: 3
Explanation: If i = 1 and j = 3:

Python solution

PythonSolution

class Solution:
    def widestPairOfIndices(self, nums1: List[int], nums2: List[int]) -> int:
        d = {0: -1}
        ans = s = 0
        for i, (a, b) in enumerate(zip(nums1, nums2)):
            s += a - b
            if s in d:
                ans = max(ans, i - d[s])
            else:
                d[s] = i
        return ans

Complexity

Time: O(n)
Space · auxiliary: O(n)where n is the length of the array nums

Pattern: Prefix Sum

Precompute running totals once so any range query becomes a single subtraction. LeetCode 1983. Widest Pair of Indices With Equal Range Sum is filed here because LeetCode tags it Prefix Sum, which is the vocabulary this hub collects.

The prefix sum guide has the Python template for the pattern and the 157 LeetCode problems that use it.

Related problems
No.	Problem	Difficulty
0523	Continuous Subarray Sum	Medium
0525	Contiguous Array	Medium
0560	Subarray Sum Equals K	Medium
0930	Binary Subarrays With Sum	Medium
0974	Subarray Sums Divisible by K	Medium
1074	Number of Submatrices That Sum to Target	Hard

Frequently asked questions

How hard is LeetCode 1983. Widest Pair of Indices With Equal Range Sum?: LeetCode 1983. Widest Pair of Indices With Equal Range Sum is rated Medium on LeetCode.
What is the time complexity of LeetCode 1983. Widest Pair of Indices With Equal Range Sum?: The Python solution on this page runs in O(n).
What is the space complexity of LeetCode 1983. Widest Pair of Indices With Equal Range Sum?: The Python solution on this page uses O(n), where n is the length of the array nums auxiliary space.
What topics does LeetCode 1983. Widest Pair of Indices With Equal Range Sum cover?: LeetCode 1983. Widest Pair of Indices With Equal Range Sum is tagged Array, Hash Table and Prefix Sum on LeetCode.
Is LeetCode 1983. Widest Pair of Indices With Equal Range Sum a premium problem?: Yes. LeetCode 1983. Widest Pair of Indices With Equal Range Sum is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.

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python

Please analyze this context and help me answer the question or solve the problem.

Transcript

"Okay, next one. Given a string, return the length of the longest substring that has no repeating characters. Walk me through your approach before you code it."

Technical interview · 45 minsolution.pyShared editor · Python 3Longest substring without repeatsMediumGiven a string s, return the length of the longest contiguous substring in which every character is unique.s = "abcabcbb"
→ 31def length_of_longest_substring(s: str) -> int:2 last_seen = {}3 left = best = 04 for right, ch in enumerate(s):5 if ch in last_seen and last_seen[ch] >= left:

python

def length_of_longest_substring(s: str) -> int:    last_seen = {}                    # char -> index    left = best = 0    for right, ch in enumerate(s):        if ch in last_seen and last_seen[ch] >= left:            left = last_seen[ch] + 1  # jump past it        last_seen[ch] = right        best = max(best, right - left + 1)    return best    last_seen = {}                    # char -> index

Approach

At each index r, the window's left edge is:

left r = max (left r -1, last[s r] + 1)

Time O(n): each character enters and leaves the window once.
Space O(min(n, k)): k is the size of the character set.

Type your message...

GPT-6.1 Terra

Transcribing...

90%Subscription

python

Please analyze this context and help me answer the question or solve the problem.

Transcript

"Okay, next one. Given a string, return the length of the longest substring that has no repeating characters. Walk me through your approach before you code it."

python

def length_of_longest_substring(s: str) -> int:    last_seen = {}                    # char -> index    left = best = 0    for right, ch in enumerate(s):        if ch in last_seen and last_seen[ch] >= left:            left = last_seen[ch] + 1  # jump past it        last_seen[ch] = right        best = max(best, right - left + 1)    return best    last_seen = {}                    # char -> index

Approach

At each index r, the window's left edge is:

left r = max (left r -1, last[s r] + 1)

Time O(n): each character enters and leaves the window once.
Space O(min(n, k)): k is the size of the character set.

Type your message...

GPT-6.1 TerraClear chat

Transcribing...

QuitEscHide+QCapture+PCrop+OSubmit+EnterPress to get answerLeft/Right+Up/Down+

90%Subscription

python

Please analyze this context and help me answer the question or solve the problem.

Transcript

"Okay, next one. Given a string, return the length of the longest substring that has no repeating characters. Walk me through your approach before you code it."

python

def length_of_longest_substring(s: str) -> int:    last_seen = {}                    # char -> index    left = best = 0    for right, ch in enumerate(s):        if ch in last_seen and last_seen[ch] >= left:            left = last_seen[ch] + 1  # jump past it        last_seen[ch] = right        best = max(best, right - left + 1)    return best    last_seen = {}                    # char -> index

Approach

At each index r, the window's left edge is:

left r = max (left r -1, last[s r] + 1)

Time O(n): each character enters and leaves the window once.
Space O(min(n, k)): k is the size of the character set.

Type your message...

GPT-6.1 TerraClear chat

Transcribing...

Primary display

QuitEscHide+QCapture+PCrop+OSubmit+EnterPress to get answerLeft/Right+Up/Down+

The problem

Example

Python solution

Complexity

Pattern: Prefix Sum

Related problems

Frequently asked questions

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Approach

Approach

Approach