# Shortest Distance to Target Color in C++

Suppose we have an array color, in which there are three colors: 1, 2 and 3. We have given some queries. Each query consists of two integers i and c, we have to find the shortest distance between the given index i and the target color c. If there is no solution, then return -1. So if the colors array is like [1,1,2,1,3,2,2,3,3], and the queries array is like [[1,3], [2,2], [6,1]], the output will be [3,0,3]. This is because the nearest 3 from index 1 is at index 4 (3 steps away). Then the nearest 2 from index 2 is at index 2 itself (0 steps away). And the nearest 1 from index 6 is at index 3 (3 steps away).

To solve this, we will follow these steps −

• Create one matrix called index with 4 rows, n := number of elements in the color array

• for I in range 0 to n – 1

• insert i into index[colors[i]]

• x := queries[i, 0] and c := queries[i, 1]

• if size of index[c] is 0, then insert -1 into ret, and skip next iteration

• it := first element that is not less than x – first element of index[c]

• op1 := infinity, op2 := infinity

• if it = size of index[c], decrease it by 1 op1 := |x – index[c, it]|

• otherwise when it = 0, then op1 := |x – index[c, it]|

• otherwise op1 := |x – index[c, it]|, decrease it by 1 and op2 := |x – index[c, it]|

• insert minimum of op1 and op2 into ret

• return ret

## Example (C++)

Let us see the following implementation to get a better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<auto> v){
cout << "[";
for(int i = 0; i<v.size(); i++){
cout << v[i] << ", ";
}
cout << "]"<<endl;
}
class Solution {
public:
vector<int> shortestDistanceColor(vector<int>& colors, vector<vector<int>>& queries) {
vector < vector <int> >idx(4);
int n = colors.size();
for(int i = 0; i < n; i++){
idx[colors[i]].push_back(i);
}
vector <int> ret;
for(int i = 0; i < queries.size(); i++){
int x = queries[i][0];
int c = queries[i][1];
if(idx[c].size() == 0){
ret.push_back(-1);
continue;
}
int it = lower_bound(idx[c].begin(), idx[c].end() , x) - idx[c].begin();
int op1 = INT_MAX;
int op2 = INT_MAX;
if(it == idx[c].size()){
it--;
op1 = abs(x - idx[c][it]);
}
else if(it == 0){
op1 = abs(x - idx[c][it]);
}
else{
op1 = abs(x - idx[c][it]);
it--;
op2 = abs(x - idx[c][it]);
}
ret.push_back(min(op1, op2));
}
return ret;
}
};
main(){
vector<int> v = {1,1,2,1,3,2,2,3,3};
vector<vector<int>> v1 = {{1,3},{2,2},{6,1}};
Solution ob;
print_vector(ob.shortestDistanceColor(v, v1));
}

## Input

[1,1,2,1,3,2,2,3,3]
[[1,3],[2,2],[6,1]]

## Output

[3,0,3]