Interval Painting - Google Top Interview Questions


Problem Statement :


You are given a list of integers walks and an integer target. 

You are currently at position 0 in a one-dimensional line. 

Each integer abs(walks[i]) represents the number of steps taken. 

Positive value means you walked right while negative value means you walked left.

We define a "block" as an interval of length 1 that has been walked on.

 For example, if you walk right 2 times, then you walked on blocks [0, 1] and [1, 2] once each. 

If you walk left once, then you'd walk on block [1, 2] again.

Return the number of blocks that's been walked on at least target number of times.

Constraints

0 ≤ n ≤ 100,000 where n is the length of walks


1 ≤ target

Example 1

Input

walks = [2, -4, 1]

target = 2

Output

3

Explanation

graph



We move right 2 steps right and then 4 steps left and then 1 step right. So we step on blocks [-2, -1], [0, 
1] and [1, 2] 2 times.


Solution :



title-img 




                        Solution in C++ :

int solve(vector<int> &nums, int target) {
    map<int, int> m;
    int current = 0;
    for (auto n : nums) {
        int last = current + n;
        if (n >= 0) {
            m[current] += 1;
            m[last] -= 1;
        } else {
            m[last] += 1;
            m[current] -= 1;
        }
        current = last;
    }
    int prev = 0;
    for (auto &[x, y] : m) {
        y += prev;
        prev = y;
    }
    int s = -1e9, ans = 0;
    for (auto &[x, y] : m) {
        if (y >= target) {
            if (s == -1e9) s = x;
        } else {
            if (s != -1e9) {
                ans += (x - s);
                s = -1e9;
            }
        }
    }
    return ans;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[] nums, int target) {
        Map<Integer, Integer> map = new TreeMap<>();

        int curr = 0;
        for (int num : nums) {
            int left = curr, right = curr;

            if (num < 0)
                left += num;
            else
                right += num;

            map.put(left, map.getOrDefault(left, 0) + 1);
            map.put(right, map.getOrDefault(right, 0) - 1);

            curr += num;
        }

        int res = 0, prev = 0, walk = 0;
        for (int key : map.keySet()) {
            if (walk >= target)
                res += Math.abs(key - prev);
            walk += map.get(key);
            prev = key;
        }

        return res;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, nums, target):
        pos = 0
        jumps = defaultdict(int)
        for dist in nums:
            jumps[pos] += 1 if dist > 0 else -1
            jumps[pos + dist] -= 1 if dist > 0 else -1
            pos += dist
        lastpos = level = total = 0
        for pos, val in sorted(jumps.items()):
            if level >= target:
                total += pos - lastpos
            level += val
            lastpos = pos
        return total


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