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Commit04d4418

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Merge pull requestneetcode-gh#3264 from adityaiiitr/adityacpp
create 1091-shortestpath-in-binary-matrix.cpp
2 parents779349c +6092475 commit04d4418

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2 files changed

+77
-1
lines changed

2 files changed

+77
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lines changed

‎cpp/0787-cheapest-flights-within-k-stops.cpp

Lines changed: 38 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -1,3 +1,40 @@
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/**
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This function uses the Bellman-Ford algorithm to find the cheapest price from source (src) to destination (dst)
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with at most k stops allowed. It iteratively relaxes the edges for k+1 iterations, updating the minimum
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cost to reach each vertex. The final result is the minimum cost to reach the destination, or -1 if the
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destination is not reachable within the given constraints.
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Space Complexity: O(n) - space used for the prices array.
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Time Complexity: O(k * |flights|) - k iterations, processing all flights in each iteration.
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*/
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classSolution {
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public:
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intfindCheapestPrice(int n, vector<vector<int>>& flights,int src,int dst,int k) {
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vector<int>prices(n, INT_MAX);
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prices[src] =0;
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// Perform k+1 iterations of Bellman-Ford algorithm.
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for (int i =0; i < k +1; i++) {
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vector<int>tmpPrices(begin(prices),end(prices));
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for (auto it : flights) {
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int s = it[0];
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int d = it[1];
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int p = it[2];
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if (prices[s] == INT_MAX)continue;
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if (prices[s] + p < tmpPrices[d]) {
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tmpPrices[d] = prices[s] + p;
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}
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}
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prices = tmpPrices;
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}
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return prices[dst] == INT_MAX ? -1 : prices[dst];
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}
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};
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/*
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Given cities connected by flights [from,to,price], also given src, dst, & k:
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Return cheapest price from src to dst with at most k stops
@@ -75,4 +112,4 @@ class Solution {
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}
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return distances[dst];
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}
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};
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};
Lines changed: 39 additions & 0 deletions
Original file line numberDiff line numberDiff line change
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classSolution {
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public:
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intshortestPathBinaryMatrix(vector<vector<int>>& grid) {
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int n = grid.size();
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if(grid[0][0]==1 || grid[n-1][n-1]==1)return -1;
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// {{r,c},length}
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queue<pair<pair<int,int>,int>> q;
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set<pair<int,int>> visited;
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q.push({{0,0},0});
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visited.insert(make_pair(0,0));
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int drow[] = {0,1,0,-1,1,-1,1,-1};
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int dcol[] = {1,0,-1,0,1,-1,-1,1};
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while(!q.empty()){
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int row = q.front().first.first;
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int col = q.front().first.second;
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int len = q.front().second;
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if(row == n-1 && col == n-1){
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return len+1;
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}
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for(int i=0; i<8; i++){
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int nrow = row + drow[i];
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int ncol = col + dcol[i];
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if(nrow>=0 && nrow<n && ncol>=0 && ncol<n && visited.find({nrow,ncol})==visited.end() && grid[nrow][ncol]==0){
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q.push({{nrow,ncol},len+1});
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visited.insert(make_pair(nrow,ncol));
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}
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}
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q.pop();
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}
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return -1;
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}
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};

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